Coordinate geometry

1. Cartesian plane

2. Straight line graphs


ID is: 1214 Seed is: 6533

Calculating a line from two points

You are given the following diagram:

Calculate the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the gradient m of the line and then the y-intercept (c). Recall that the the equation of a straight line is of the form y=mx+c.


STEP: Calculate the gradient m of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(2.5)(2.5)(1)(4)m=1

STEP: Calculate the y-intercept (c) of the line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituing a point A into the general form for a straight line:

y=mx+c(2.5)=(1)×(4)+cc=1.5

STEP: Substitute the values of m and c into the equation y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1x+1.5

Submit your answer as:

ID is: 1214 Seed is: 8200

Calculating a line from two points

You are given the following diagram:

Calculate the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the gradient m of the line and then the y-intercept (c). Recall that the the equation of a straight line is of the form y=mx+c.


STEP: Calculate the gradient m of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(2)(4)(2)(1)m=2

STEP: Calculate the y-intercept (c) of the line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituing a point A into the general form for a straight line:

y=mx+c(4)=(2)×(1)+cc=2

STEP: Substitute the values of m and c into the equation y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=2x+2

Submit your answer as:

ID is: 1214 Seed is: 1039

Calculating a line from two points

You are given the following diagram:

Calculate the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the gradient m of the line and then the y-intercept (c). Recall that the the equation of a straight line is of the form y=mx+c.


STEP: Calculate the gradient m of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(0.5)(3.5)(1)(2)m=1

STEP: Calculate the y-intercept (c) of the line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituing a point A into the general form for a straight line:

y=mx+c(3.5)=(1)×(2)+cc=1.5

STEP: Substitute the values of m and c into the equation y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1x1.5

Submit your answer as:

ID is: 2678 Seed is: 140

Determining the equation of a straight line

In the diagram below, the line passes through the point A(2;3) and has a gradient m=2.

Find the equation of the line.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of the equation of a straight line:

y=mx+c

In the question statement we are given m, so the only unknown is c.

y=(2)x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

We need to calculate the value of the y-intercept (c) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

y=(2)x+c3=(2)(2)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the straight line is:

y=2x1

Submit your answer as:

ID is: 2678 Seed is: 7982

Determining the equation of a straight line

In the diagram below, the line passes through the point A(5;1.5) and has a gradient m=0.5.

Determine the equation of the line.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of the equation of a straight line:

y=mx+c

In the question statement we are given m, so the only unknown is c.

y=(0.5)x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

We need to calculate the value of the y-intercept (c) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

y=(0.5)x+c1.5=(0.5)(5)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the straight line is:

y=0.5x+1

Submit your answer as:

ID is: 2678 Seed is: 2658

Determining the equation of a straight line

In the diagram below, the line passes through the point A(1;1) and has a gradient m=0.5.

Calculate the equation of the line.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of the equation of a straight line:

y=mx+c

In the question statement we are given m, so the only unknown is c.

y=(0.5)x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

We need to calculate the value of the y-intercept (c) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

y=(0.5)x+c1=(0.5)(1)+cc=1.5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the straight line is:

y=0.5x+1.5

Submit your answer as:

ID is: 2686 Seed is: 7942

The gradient-intercept form of the straight line equation

Given the following diagram of a straight line passing through the points A(1;1.5) and B(3;2.5).

Determine the equation of the line AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient m of the line:

m=yByAxBxA=(2.5)(1.5)(3)(1)=1

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c). We do this by substituing the coordinates of point A into the general form for a straight line:

y=mx+c1.5=(1)×(1)+cc=0.5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is

y=x+0.5

Submit your answer as:

ID is: 2686 Seed is: 4420

The gradient-intercept form of the straight line equation

Given the following diagram of a straight line passing through the points A(4;2.5) and B(1;2.5).

Calculate the equation of the line AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient m of the line:

m=yByAxBxA=(2.5)(2.5)(1)(4)=1

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c). We do this by substituing the coordinates of point A into the general form for a straight line:

y=mx+c2.5=(1)×(4)+cc=1.5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is

y=x1.5

Submit your answer as:

ID is: 2686 Seed is: 9816

The gradient-intercept form of the straight line equation

Given the following diagram of a straight line passing through the points A(1;3.5) and B(1;0.5).

Find the equation of the line AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient m of the line:

m=yByAxBxA=(0.5)(3.5)(1)(1)=2

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c). We do this by substituing the coordinates of point A into the general form for a straight line:

y=mx+c3.5=(2)×(1)+cc=1.5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is

y=2x+1.5

Submit your answer as:

ID is: 2790 Seed is: 2632

Finding the equations of parallel lines

Consider the diagram given below:

Line AB is parallel to y=1,5x+5 (the green line).

Determine the equation of AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of AB, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

m1=m2=1.5General equation:y=mx+cy=1.5x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we calculate the value of the c by substituting the coordinates of point A into the equation:

y=1.5x+c2.5=(1.5)(3)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is

y=1.5x+2

Submit your answer as:

ID is: 2790 Seed is: 7516

Finding the equations of parallel lines

Consider the diagram given below:

Line AB is parallel to y=1,5x+5,5 (the green line).

Find the equation of AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of AB, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

m1=m2=1.5General equation:y=mx+cy=1.5x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we calculate the value of the c by substituting the coordinates of point A into the equation:

y=1.5x+c0=(1.5)(1)+cc=1.5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is

y=1.5x+1.5

Submit your answer as:

ID is: 2790 Seed is: 1871

Finding the equations of parallel lines

Consider the diagram given below:

Line AB is parallel to y=1,5x+1 (the green line).

Determine the equation of AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of AB, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

m1=m2=1.5General equation:y=mx+cy=1.5x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we calculate the value of the c by substituting the coordinates of point A into the equation:

y=1.5x+c3.5=(1.5)(1)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is

y=1.5x2

Submit your answer as:

ID is: 1218 Seed is: 1252

Calculating a gradient from two points

You are given the diagram below. The coordinates are A:(2;4) and B:(1;0.5).

Calculate the gradient (m) of line AB, correct to 2 decimal places.

Answer: Gradient (m) = .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Calculate the gradient m using the formula:

m=yByAxBxA,

where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the gradient formula
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is m):

m=y2y1x2x1

STEP: Substitute the coordinates of points A and B into the gradient formula
[−1 point ⇒ 0 / 2 points left]

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

m=(0.5)(4)(1)(2)m=4.53m=1.5

Therefore the gradient m of the line AB is 1.5.


Submit your answer as:

ID is: 1218 Seed is: 3000

Calculating a gradient from two points

You are given the diagram below. The coordinates are A:(2;3) and B:(1;0).

Calculate the gradient (m) of line AB, correct to 2 decimal places.

Answer: Gradient (m) = .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Calculate the gradient m using the formula:

m=yByAxBxA,

where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the gradient formula
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is m):

m=y2y1x2x1

STEP: Substitute the coordinates of points A and B into the gradient formula
[−1 point ⇒ 0 / 2 points left]

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

m=(0)(3)(1)(2)m=33m=1

Therefore the gradient m of the line AB is 1.


Submit your answer as:

ID is: 1218 Seed is: 4922

Calculating a gradient from two points

You are given the diagram below. The coordinates are A:(3;4) and B:(5;4).

Calculate the gradient (m) of line AB, correct to 2 decimal places.

Answer: Gradient (m) = .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Calculate the gradient m using the formula:

m=yByAxBxA,

where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the gradient formula
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is m):

m=y2y1x2x1

STEP: Substitute the coordinates of points A and B into the gradient formula
[−1 point ⇒ 0 / 2 points left]

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

m=(4)(4)(5)(3)m=88m=1

Therefore the gradient m of the line AB is 1.


Submit your answer as:

ID is: 1299 Seed is: 1004

Finding an unknown coordinate on a line

Consider the diagram below, which shows a straight line. The line passes through the points A(3;112) and B(x;52).

The equation of the line is: y=4x3+32.

Determine the x-coordinate of Point B.

INSTRUCTION: Your answer should be exact (do not round off).
Answer: The x-coordinate of Point B is: .
numeric
2 attempts remaining
STEP: Use the equation of the line
[−2 points ⇒ 0 / 2 points left]

We have a line passing through two points, but Point B has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute B(x;52) - the coordinates of B - into the equation and solve for the unknown value. To make it a bit more clear, we can use xB to show that the coordinate we want belongs to Point B.

y=4x3+32(52)=4xB3+324=4xB33=xB
NOTE: We did not need the coordinates of Point A for any part of this solution. Since we already know the equation of the line, we can use Point B immediately in the equation.

The x-coordinate of Point B is 3.


Submit your answer as:

ID is: 1299 Seed is: 1369

Finding an unknown coordinate on a line

Consider the diagram below, which shows a straight line. The line passes through the points A(6;132) and B(x;12).

The equation of the line is: y=2x352.

Determine the x-coordinate of Point B.

INSTRUCTION: Your answer should be exact (do not round off).
Answer: The x-coordinate of Point B is: .
numeric
2 attempts remaining
STEP: Use the equation of the line
[−2 points ⇒ 0 / 2 points left]

We have a line passing through two points, but Point B has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute B(x;12) - the coordinates of B - into the equation and solve for the unknown value. To make it a bit more clear, we can use xB to show that the coordinate we want belongs to Point B.

y=2x352(12)=2xB3522=2xB33=xB
NOTE: We did not need the coordinates of Point A for any part of this solution. Since we already know the equation of the line, we can use Point B immediately in the equation.

The x-coordinate of Point B is 3.


Submit your answer as:

ID is: 1299 Seed is: 3803

Finding an unknown coordinate on a line

Consider the diagram below, which shows a straight line. The line passes through the points A(3;7) and B(2;y).

The equation of the line is: y=5x212.

Find the y-coordinate of Point B.

INSTRUCTION: Your answer should be exact (do not round off).
Answer: The y-coordinate of Point B is: .
numeric
2 attempts remaining
STEP: Use the equation of the line
[−2 points ⇒ 0 / 2 points left]

We have a line passing through two points, but Point B has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute B(2;y) - the coordinates of B - into the equation and solve for the unknown value. To make it a bit more clear, we can use yB to show that the coordinate we want belongs to Point B.

y=5x212yB=5(2)212yB=112
NOTE: We did not need the coordinates of Point A for any part of this solution. Since we already know the equation of the line, we can use Point B immediately in the equation.

The y-coordinate of Point B is 112.


Submit your answer as:

ID is: 1211 Seed is: 4318

Finding the coordinate of a point using parallel lines

The following diagram shows two parallel lines. Point A is at (1;0) and Point B is at (x;4). Line AB runs parallel to the following line: y=2x5, which is the dashed blue line.

Determine the x-coordinate of Point B.

Answer: Point B = (;-4).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the fact that parallel lines have equal gradients.


STEP: Start building the equation for line AB
[−1 point ⇒ 2 / 3 points left]

We need to find the x-coordinate of Point B. We do not know much about line AB. But we do know that it is parallel to the line y=2x5. That means the lines have the same gradient! We can read the gradient from the equation.

y=2x5gradient=2

Line AB has the same gradient, so we can start building the equation for the line:

y=mx+cyAB=2x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

We know the coordinates of Point A. We can use them with this equation to calculate the value of c.

y=2x+c(0)=2(1)+cc=2

Great! Now we have the equation for line AB:yAB=2x2.


STEP: Substitute the coordinates of Point B into the equation and solve for xB
[−1 point ⇒ 0 / 3 points left]

Finally, we substitute the coordinates of Point B into the equation for line AB:

y=2x2(4)=2(xB)2

Solving, we get:

xB=1
NOTE:

You could also solve this question using the gradient formula. The gradient of line AB is −2 so we could set up this equation using the coordinates from Points A and B:

m=y2y1x2x12=40xB(1)

If you solve this equation, you will get the same answer that we got using the equation of line AB.

The x-coordinate of Point B is 1.


Submit your answer as:

ID is: 1211 Seed is: 582

Finding the coordinate of a point using parallel lines

The following diagram shows two parallel lines. Point A is at (1;3) and Point B is at (x;1). Line AB runs parallel to the following line: y=x3, which is the dashed blue line.

Determine the x-coordinate of Point B.

Answer: Point B = (;1).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the fact that parallel lines have equal gradients.


STEP: Start building the equation for line AB
[−1 point ⇒ 2 / 3 points left]

We need to find the x-coordinate of Point B. We do not know much about line AB. But we do know that it is parallel to the line y=x3. That means the lines have the same gradient! We can read the gradient from the equation.

y=x3gradient=1

Line AB has the same gradient, so we can start building the equation for the line:

y=mx+cyAB=1x+c=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

We know the coordinates of Point A. We can use them with this equation to calculate the value of c.

y=x+c(3)=1(1)+cc=2

Great! Now we have the equation for line AB:yAB=x+2.


STEP: Substitute the coordinates of Point B into the equation and solve for xB
[−1 point ⇒ 0 / 3 points left]

Finally, we substitute the coordinates of Point B into the equation for line AB:

y=x+2(1)=(xB)+2

Solving, we get:

xB=1
NOTE:

You could also solve this question using the gradient formula. The gradient of line AB is −1 so we could set up this equation using the coordinates from Points A and B:

m=y2y1x2x11=13xB(1)

If you solve this equation, you will get the same answer that we got using the equation of line AB.

The x-coordinate of Point B is 1.


Submit your answer as:

ID is: 1211 Seed is: 8720

Finding the coordinate of a point using parallel lines

The following diagram shows two parallel lines. Point A is at (1;52) and Point B is at (x;32). Line AB runs parallel to the following line: y=2x72, which is the dashed blue line.

Determine the x-coordinate of Point B.

Answer: Point B = (;- \frac{3}{2}).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the fact that parallel lines have equal gradients.


STEP: Start building the equation for line AB
[−1 point ⇒ 2 / 3 points left]

We need to find the x-coordinate of Point B. We do not know much about line AB. But we do know that it is parallel to the line y=2x72. That means the lines have the same gradient! We can read the gradient from the equation.

y=2x72gradient=2

Line AB has the same gradient, so we can start building the equation for the line:

y=mx+cyAB=2x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

We know the coordinates of Point A. We can use them with this equation to calculate the value of c.

y=2x+c(52)=2(1)+cc=12

Great! Now we have the equation for line AB:yAB=2x+12.


STEP: Substitute the coordinates of Point B into the equation and solve for xB
[−1 point ⇒ 0 / 3 points left]

Finally, we substitute the coordinates of Point B into the equation for line AB:

y=2x+12(32)=2(xB)+12

Solving, we get:

xB=1
NOTE:

You could also solve this question using the gradient formula. The gradient of line AB is −2 so we could set up this equation using the coordinates from Points A and B:

m=y2y1x2x12=3252xB(1)

If you solve this equation, you will get the same answer that we got using the equation of line AB.

The x-coordinate of Point B is 1.


Submit your answer as:

ID is: 2763 Seed is: 6009

Working with parallel lines

The diagram below shows two parallel lines. The dashed red line has the equation:

ydashed=3x2+7

The solid blue line passes through the points A(6;7) and B(1;y), as shown. Note that the y-coordinate of point B is unknown.

Determine yB, the y-coordinate of point B.

Answer: yB=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line AB by looking at the equation for the other line.


STEP: Use the fact that parallel lines have equal gradients
[−1 point ⇒ 3 / 4 points left]

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

y=3x2+7

which means that both of the lines have a gradient of 32. So the equation of line AB must be:

y=3x2+c

STEP: Find the value of c for the line using point A
[−1 point ⇒ 2 / 4 points left]

In this question we need a missing coordinate from point B. However, we know both coordinates for point A, so we can use them to find the value of c in the equation for line AB.

y=3x2+c(7)=3(6)2+cof point A to find cUse the coordinates2=c

Now we have the complete equation for line AB:

y=3x2+2

STEP: Use the x-coordinate of point B to find yB
[−2 points ⇒ 0 / 4 points left]

Now we can substitute the x-coordinate for point B into the equation for line AB and solve for yB. This works because the equation of any line agrees with all of the points on the line: if we substitute in the x-coordinate then when we solve for y we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

yB=3xB2+2yB=3(1)2+2yB=72

The correct answer is yB=72.


Submit your answer as:

ID is: 2763 Seed is: 1748

Working with parallel lines

The diagram below shows two parallel lines. The dashed red line has the equation:

ydashed=x252

The solid blue line passes through the points A(4;12) and B(8;y), as shown. Note that the y-coordinate of point B is unknown.

Find yB, the y-coordinate of point B.

Answer: yB=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line AB by looking at the equation for the other line.


STEP: Use the fact that parallel lines have equal gradients
[−1 point ⇒ 3 / 4 points left]

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

y=x252

which means that both of the lines have a gradient of 12. So the equation of line AB must be:

y=x2+c

STEP: Find the value of c for the line using point A
[−1 point ⇒ 2 / 4 points left]

In this question we need a missing coordinate from point B. However, we know both coordinates for point A, so we can use them to find the value of c in the equation for line AB.

y=x2+c(12)=(4)2+cof point A to find cUse the coordinates52=c

Now we have the complete equation for line AB:

y=x2+52

STEP: Use the x-coordinate of point B to find yB
[−2 points ⇒ 0 / 4 points left]

Now we can substitute the x-coordinate for point B into the equation for line AB and solve for yB. This works because the equation of any line agrees with all of the points on the line: if we substitute in the x-coordinate then when we solve for y we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

yB=xB2+52yB=(8)2+52yB=132

The correct answer is yB=132.


Submit your answer as:

ID is: 2763 Seed is: 9169

Working with parallel lines

The diagram below shows two parallel lines. The dashed red line has the equation:

ydashed=2x3+4

The solid blue line passes through the points A(x;133) and B(5;73), as shown. Note that the x-coordinate of point A is unknown.

Find xA, the x-coordinate of point A.

Answer: xA=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line AB by looking at the equation for the other line.


STEP: Use the fact that parallel lines have equal gradients
[−1 point ⇒ 3 / 4 points left]

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

y=2x3+4

which means that both of the lines have a gradient of 23. So the equation of line AB must be:

y=2x3+c

STEP: Find the value of c for the line using point B
[−1 point ⇒ 2 / 4 points left]

In this question we need a missing coordinate from point A. However, we know both coordinates for point B, so we can use them to find the value of c in the equation for line AB.

y=2x3+c(73)=2(5)3+cof point B to find cUse the coordinates1=c

Now we have the complete equation for line AB:

y=2x31

STEP: Use the y-coordinate of point A to find xA
[−2 points ⇒ 0 / 4 points left]

Now we can substitute the y-coordinate for point A into the equation for line AB and solve for xA. This works because the equation of any line agrees with all of the points on the line: if we substitute in the y-coordinate then when we solve for x we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

yA=2xA31(133)=2xA31xA=5

The correct answer is xA=5.


Submit your answer as:

ID is: 1297 Seed is: 9286

Finding the equation of a straight line

The diagram below shows a line. The line has a gradient m=12 and it passes through the point A(7;5).

Find the equation of the line. Give only the right side of the equation, following "y= " below.

Answer: Equation of the line: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should start with the standard form for a straight line (y=mx+c) and substitute in the gradient. Then use the coordinates of point A to find c.


STEP: Start building the equation using the gradient
[−1 point ⇒ 1 / 2 points left]

The first thing to remember is that a straight line has this general form:

y=mx+c

where m is the gradient and c is the y-intercept. We have the gradient m of the line (it is given in the question) so we can substitute that into the equation of the line:

y=(12)x+cy=x2+c

STEP: Use the coordinates of the point to find c
[−1 point ⇒ 0 / 2 points left]

Now we need the value of c for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point A we will use them.

y=x2+c(5)=(7)2+c5=72+c32=c

Notice that c is negative: this agrees with the diagram, because the y-intercept is below the x-axis.

The equation of the line is:

y=x232

Submit your answer as:

ID is: 1297 Seed is: 3411

Finding the equation of a straight line

The diagram below shows a line. The line has a gradient m=23 and it passes through the point A(3;1).

Calculate the equation of the line. Give only the right side of the equation, following "y= " below.

Answer: Equation of the line: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should start with the standard form for a straight line (y=mx+c) and substitute in the gradient. Then use the coordinates of point A to find c.


STEP: Start building the equation using the gradient
[−1 point ⇒ 1 / 2 points left]

The first thing to remember is that a straight line has this general form:

y=mx+c

where m is the gradient and c is the y-intercept. We have the gradient m of the line (it is given in the question) so we can substitute that into the equation of the line:

y=(23)x+cy=2x3+c

STEP: Use the coordinates of the point to find c
[−1 point ⇒ 0 / 2 points left]

Now we need the value of c for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point A we will use them.

y=2x3+c(1)=2(3)3+c1=2+c1=c

Notice that c is negative: this agrees with the diagram, because the y-intercept is below the x-axis.

The equation of the line is:

y=2x31

Submit your answer as:

ID is: 1297 Seed is: 7888

Finding the equation of a straight line

The diagram below shows a line. The line has a gradient m=23 and it passes through the point A(6;52).

Determine the equation of the line. Give only the right side of the equation, following "y= " below.

Answer: Equation of the line: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should start with the standard form for a straight line (y=mx+c) and substitute in the gradient. Then use the coordinates of point A to find c.


STEP: Start building the equation using the gradient
[−1 point ⇒ 1 / 2 points left]

The first thing to remember is that a straight line has this general form:

y=mx+c

where m is the gradient and c is the y-intercept. We have the gradient m of the line (it is given in the question) so we can substitute that into the equation of the line:

y=(23)x+cy=2x3+c

STEP: Use the coordinates of the point to find c
[−1 point ⇒ 0 / 2 points left]

Now we need the value of c for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point A we will use them.

y=2x3+c(52)=2(6)3+c52=4+c32=c

Notice that c is negative: this agrees with the diagram, because the y-intercept is below the x-axis.

The equation of the line is:

y=2x332

Submit your answer as:

ID is: 1209 Seed is: 9587

Finding a point on a line

You are given the following diagram:

Point C lies on the line AB.

Calculate the missing coordinate of Point C.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point C = (;0.4).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the equation of the straight line AB.


STEP: Calculate the gradient (m) of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(0.5)(2.5)(5)(1)m=0.5

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting Point A into the general form for a straight line:

y=mx+c(2.5)=(0.5)×(1)+cc=2

STEP: Solve for x using the known value of Point C
[−1 point ⇒ 0 / 3 points left]

We now substitute the known values of Point C into the known equation of the straight line:

y=mx+cy=(0.5)x20.4=(0.5)(x)2x=3.2

Submit your answer as:

ID is: 1209 Seed is: 3304

Finding a point on a line

You are given the following diagram:

Point C lies on the line AB.

Calculate the missing coordinate of Point C.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point C = (0.3;).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the equation of the straight line AB.


STEP: Calculate the gradient (m) of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(3.5)(0.5)(1)(1)m=2

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting Point A into the general form for a straight line:

y=mx+c(0.5)=(2)×(1)+cc=1.5

STEP: Solve for y using the known value of Point C
[−1 point ⇒ 0 / 3 points left]

We now substitute the known values of Point C into the known equation of the straight line:

y=mx+cy=(2)x1.5y=(2)(0.3)1.5y=0.9

Submit your answer as:

ID is: 1209 Seed is: 8369

Finding a point on a line

You are given the following diagram:

Point C lies on the line AB.

Calculate the missing coordinate of Point C.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point C = (;0.3).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the equation of the straight line AB.


STEP: Calculate the gradient (m) of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(2)(3)(5)(5)m=0.5

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting Point A into the general form for a straight line:

y=mx+c(3)=(0.5)×(5)+cc=0.5

STEP: Solve for x using the known value of Point C
[−1 point ⇒ 0 / 3 points left]

We now substitute the known values of Point C into the known equation of the straight line:

y=mx+cy=(0.5)x+0.50.3=(0.5)(x)+0.5x=1.6

Submit your answer as:

ID is: 1207 Seed is: 8604

Finding a coordinate of a point using an equation

You are given the following diagram. It shows point A at (−2;4) and point B at (3;y).

The equation of line segment AB is: y=1.5x+1

Determine the y coordinate of point B, correct to 2 decimal places.

Answer: Point B = (3;)
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the equation together with the coordinates of point B.


STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 2 / 3 points left]

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

y=−1.5x+1

Every point on the line must agree with this equation. That includes the coordinates of point B. We substitute the known value for point B into the equation because that will give us the value of the missing coordinate y which agrees with the equation.

y=1.5x+1y=1.5(3)+1

STEP: Solve the equation for the value of y
[−2 points ⇒ 0 / 3 points left]

Solving, we get:

y=3.5

Submit your answer as:

ID is: 1207 Seed is: 2557

Finding a coordinate of a point using an equation

You are given the following diagram. It shows point A at (−1;2.5) and point B at (1;y).

The equation of line segment AB is: y=2x+0.5

Calculate the y coordinate of point B, correct to 2 decimal places.

Answer: Point B = (1;)
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the equation together with the coordinates of point B.


STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 2 / 3 points left]

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

y=−2x+0.5

Every point on the line must agree with this equation. That includes the coordinates of point B. We substitute the known value for point B into the equation because that will give us the value of the missing coordinate y which agrees with the equation.

y=2x+0.5y=2(1)+0.5

STEP: Solve the equation for the value of y
[−2 points ⇒ 0 / 3 points left]

Solving, we get:

y=1.5

Submit your answer as:

ID is: 1207 Seed is: 4024

Finding a coordinate of a point using an equation

You are given the following diagram. It shows point A at (−4;2) and point B at (x;3).

The equation of line segment AB is: y=x2

Calculate the x coordinate of point B, correct to 2 decimal places.

Answer: Point B = (;3)
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the equation together with the coordinates of point B.


STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 2 / 3 points left]

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

y=x−2

Every point on the line must agree with this equation. That includes the coordinates of point B. We substitute the known value for point B into the equation because that will give us the value of the missing coordinate x which agrees with the equation.

y=x2(3)=x2

STEP: Solve the equation for the value of x
[−2 points ⇒ 0 / 3 points left]

Solving, we get:

x=1

Submit your answer as:

ID is: 2811 Seed is: 120

Finding the equation of a parallel line

What is the equation of a line which is parallel to the line y=3x2+1 and passes through the point A(7;5)?

INSTRUCTION: Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.
Answer:
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: y=3x2+1. The gradient for this line is m=32.

For a line which is parallel to this one, the gradient of the line we want is also m=32.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now use the point (7;5), given together with the gradient, to get the equation of the line. Substitute the x and y-values into the equation to get the value of c:

y=mx+c5=(32)(7)+c5=212+c312=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=3x2312.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point A(7;5) is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.


Submit your answer as:

ID is: 2811 Seed is: 8086

Finding the equation of a parallel line

Given the equation y=2x1. Find the equation of a line which passes through P(4;4) and is parallel to the given line.

INSTRUCTION: Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.
Answer:
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: y=2x1. The gradient for this line is m=2.

For a line which is parallel to this one, the gradient of the line we want is also m=2.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now use the point (4;4), given together with the gradient, to get the equation of the line. Substitute the x and y-values into the equation to get the value of c:

y=mx+c4=(2)(4)+c4=8+c4=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=2x+4.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point P(4;4) is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.


Submit your answer as:

ID is: 2811 Seed is: 61

Finding the equation of a parallel line

What is the equation of a line which is parallel to the line y=2x2 and passes through the point N(4;0)?

INSTRUCTION: Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.
Answer:
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: y=2x2. The gradient for this line is m=2.

For a line which is parallel to this one, the gradient of the line we want is also m=2.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now use the point (4;0), given together with the gradient, to get the equation of the line. Substitute the x and y-values into the equation to get the value of c:

y=mx+c0=(2)(4)+c0=8+c8=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=2x8.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point N(4;0) is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.


Submit your answer as:

ID is: 1300 Seed is: 6832

Finding the equation of a straight line

The diagram below shows a straight line. The line passes through the coordinates A(6;112) and B(2;12). Find the equation of the line AB.

INSTRUCTION: Type the right side of the equation in the form mx+c.
Answer: y =
expression
2 attempts remaining
STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

m=yByAxBxA=(12)(112)(2)(6)=34

So the gradient of the line is 34.


STEP: Find the y-intercept of the line
[−1 point ⇒ 0 / 2 points left]

Now we need to find the y-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point A. (You can use point B if you want: any point on the line will work.)

y=mx+c(112)=(34)(6)+c112=92+c1=c

Therefore, the equation of the line AB is as follows:

y=3x41

Submit your answer as:

ID is: 1300 Seed is: 4762

Finding the equation of a straight line

The diagram below shows a straight line. The line passes through the coordinates A(1;5) and B(1;1). Determine the equation of the line AB.

INSTRUCTION: Type the right side of the equation in the form mx+c.
Answer: y =
expression
2 attempts remaining
STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

m=yByAxBxA=(1)(5)(1)(1)=2

So the gradient of the line is 2.


STEP: Find the y-intercept of the line
[−1 point ⇒ 0 / 2 points left]

Now we need to find the y-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point A. (You can use point B if you want: any point on the line will work.)

y=mx+c(5)=(2)(1)+c5=2+c3=c

Therefore, the equation of the line AB is as follows:

y=2x+3

Submit your answer as:

ID is: 1300 Seed is: 5075

Finding the equation of a straight line

The diagram below shows a straight line. The line passes through the coordinates A(1;72) and B(2;52). Find the equation of the line AB.

INSTRUCTION: Type the right side of the equation in the form mx+c.
Answer: y =
expression
2 attempts remaining
STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

m=yByAxBxA=(52)(72)(2)(1)=2

So the gradient of the line is 2.


STEP: Find the y-intercept of the line
[−1 point ⇒ 0 / 2 points left]

Now we need to find the y-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point A. (You can use point B if you want: any point on the line will work.)

y=mx+c(72)=(2)(1)+c72=2+c32=c

Therefore, the equation of the line AB is as follows:

y=2x32

Submit your answer as:

ID is: 2682 Seed is: 6328

Calculating the coordinates of a point on a line

Points A(1;2,5) and B(x;1,5) lie on the straight line y=x+1,5.

Calculate the x-coordinate of point B, rounded to 2 decimal places.

Answer: Point B = (;1.5)
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

We are given the equation of the straight line:

y=x+1,5

To determine the value of x, we substitute the y-coordinate for point B into the equation:

y=1x+1.51.5=(1)(x)+1.5

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We rearrange the equation to get the answer:

x=3

Submit your answer as:

ID is: 2682 Seed is: 6574

Calculating the coordinates of a point on a line

Points A(1;0,5) and B(3;y) lie on the straight line y=0,5x1.

Find the y-coordinate of point B, rounded to 2 decimal places.

Answer: Point B = (3;)
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

We are given the equation of the straight line:

y=0,5x1

To determine the value of y, we substitute the x-coordinate for point B into the equation:

y=0.5x1y=(0.5)(3)1

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We rearrange the equation to get the answer:

y=2.5

Submit your answer as:

ID is: 2682 Seed is: 2134

Calculating the coordinates of a point on a line

Points A(1;0,5) and B(x;2,5) lie on the straight line y=1,5x+1.

Determine the x-coordinate of point B, rounded to 2 decimal places.

Answer: Point B = (;2.5)
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

We are given the equation of the straight line:

y=1,5x+1

To determine the value of x, we substitute the y-coordinate for point B into the equation:

y=1.5x+12.5=(1.5)(x)+1

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We rearrange the equation to get the answer:

x=1

Submit your answer as:

ID is: 2676 Seed is: 5888

Identifying parallel lines

Are the following lines parallel?

3y=43x=1

Note: In tests and exams you must show calculations to support your answer.

Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, you need to know the gradient of each equation. That means that must start by arranging the equations to be in standard form: y=mx+c. If the equations do not contain the variable y, then you need to isolate x to get an equation which says, "x= a number."
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

3y=413(3y)=13(4)y=43

This equation is the same as y=0x+43. Thus the gradient of this equation is m=0.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now change the second equation to standard form. This is an unusual case: there is no y in the equation! You need to isolate the x.

3x=113(3x)=13(1)x=13

This equation does not show the gradient because it is not in standard form! You need to remember that an equation like this one (which says x= a number) represents a vertical line; and that means the gradient is undefined.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Compare the two gradients to decide if the lines are parallel or not. This is a special situation: one of the lines has a gradient of zero, while the other gradient is undefined. Remember that a horizontal line has a zero gradient and for a vertical line the gradient is undefined. This means that these two lines are perpendicular to each other. Therefore, the correct response is: not parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that there is one horizontal line and one vertical line. In fact, these lines are perpendicular to each other.


Submit your answer as:

ID is: 2676 Seed is: 6678

Identifying parallel lines

Are the following lines parallel?

2y=13x=2

Note: In tests and exams you must show calculations to support your answer.

Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, you need to know the gradient of each equation. That means that must start by arranging the equations to be in standard form: y=mx+c. If the equations do not contain the variable y, then you need to isolate x to get an equation which says, "x= a number."
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

2y=112(2y)=12(1)y=12

This equation is the same as y=0x12. Thus the gradient of this equation is m=0.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now change the second equation to standard form. This is an unusual case: there is no y in the equation! You need to isolate the x.

3x=213(3x)=13(2)x=23

This equation does not show the gradient because it is not in standard form! You need to remember that an equation like this one (which says x= a number) represents a vertical line; and that means the gradient is undefined.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Compare the two gradients to decide if the lines are parallel or not. This is a special situation: one of the lines has a gradient of zero, while the other gradient is undefined. Remember that a horizontal line has a zero gradient and for a vertical line the gradient is undefined. This means that these two lines are perpendicular to each other. Therefore, the correct response is: not parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that there is one horizontal line and one vertical line. In fact, these lines are perpendicular to each other.


Submit your answer as:

ID is: 2676 Seed is: 3026

Identifying parallel lines

Determine whether or not the following lines are parallel.

16y=20x+4y+153y+4=5x2

Note: In tests and exams you must show calculations to support your answer.

Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, you need to know the gradient of each equation. That means that must start by arranging the equations to be in standard form: y=mx+c. If the equations do not contain the variable y, then you need to isolate x to get an equation which says, "x= a number."
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

16y=20x+4y+1512y=20x+15112(12y)=112(20x+15)y=5x3+54

The gradient of the first equation is m=53.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now change the second equation to standard form.

3y+4=5x23y=5x613(3y)=13(5x6)y=5x32

The gradient of the second equation is m=53.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Compare the two gradients to decide if the lines are parallel or not. The equations have equal gradients, so they must be parallel lines. The correct choice is: parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that the two lines run parallel to one another.


Submit your answer as:

ID is: 1213 Seed is: 9644

Calculating a coordinate for a point using the gradient

Look at the following diagram. Point A is at (−4;−3.5) and Point B is at (x;1). Line AB has a gradient (m) of 0.5.

Calculate the x-coordinate of Point B.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point B= (;-1)
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.


STEP: Find the equation of the line segment
[−2 points ⇒ 2 / 4 points left]

We have two endpoints of a line segment, and the gradient of the segment. However, the x-coordinate of Point B is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

y=mx+c

Firstly, we can calculate the value of the y-intercept (c) of the line AB. We do this by substituting the gradient value and the coordinates of Point A into the general form for a straight line:

y=mx+c(3.5)=(0.5)×(4)+cc=1.5

This means that the equation for the line segment in the graph is:

y=0.5x1.5

STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known value for Point B into the equation:

y=0.5x1.5(1)=0.5x1.5

STEP: Solve the equation for x
[−1 point ⇒ 0 / 4 points left]

Solving, we get:

x=1

Submit your answer as:

ID is: 1213 Seed is: 5969

Calculating a coordinate for a point using the gradient

Look at the following diagram. Point A is at (−3;4) and Point B is at (x;4). Line AB has a gradient (m) of −2.

Calculate the x-coordinate of Point B.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point B= (;-4)
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.


STEP: Find the equation of the line segment
[−2 points ⇒ 2 / 4 points left]

We have two endpoints of a line segment, and the gradient of the segment. However, the x-coordinate of Point B is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

y=mx+c

Firstly, we can calculate the value of the y-intercept (c) of the line AB. We do this by substituting the gradient value and the coordinates of Point A into the general form for a straight line:

y=mx+c(4)=(2)×(3)+cc=2

This means that the equation for the line segment in the graph is:

y=2x2

STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known value for Point B into the equation:

y=2x2(4)=2x2

STEP: Solve the equation for x
[−1 point ⇒ 0 / 4 points left]

Solving, we get:

x=1

Submit your answer as:

ID is: 1213 Seed is: 8368

Calculating a coordinate for a point using the gradient

Look at the following diagram. Point A is at (−1;0) and Point B is at (1;y). Line AB has a gradient (m) of 2.

Calculate the y-coordinate of Point B.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point B= (1;)
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.


STEP: Find the equation of the line segment
[−2 points ⇒ 2 / 4 points left]

We have two endpoints of a line segment, and the gradient of the segment. However, the y-coordinate of Point B is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

y=mx+c

Firstly, we can calculate the value of the y-intercept (c) of the line AB. We do this by substituting the gradient value and the coordinates of Point A into the general form for a straight line:

y=mx+c(0)=(2)×(1)+cc=2

This means that the equation for the line segment in the graph is:

y=2x+2

STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known value for Point B into the equation:

y=2x+2y=2(1)+2

STEP: Solve the equation for y
[−1 point ⇒ 0 / 4 points left]

Solving, we get:

y=4

Submit your answer as:

ID is: 1217 Seed is: 400

Finding an equation with parallel lines

You are given the following diagram:

You are also told that line segment AB runs parallel to the following line: y=1x+3. Point A is at (3;2). Work out the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by recalling that the gradient of parallel lines is equal.


STEP: Determine the gradient of line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of the line AB. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line AB:

y=mx+cy=(1)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(2)=(1)×(3)+cc=1

STEP: Write down the equation of the line AB in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1x1

Submit your answer as:

ID is: 1217 Seed is: 8909

Finding an equation with parallel lines

You are given the following diagram:

You are also told that line segment AB runs parallel to the following line: y=1x+1. Point A is at (3;1). Work out the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by recalling that the gradient of parallel lines is equal.


STEP: Determine the gradient of line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of the line AB. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line AB:

y=mx+cy=(1)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(1)=(1)×(3)+cc=2

STEP: Write down the equation of the line AB in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1x2

Submit your answer as:

ID is: 1217 Seed is: 1866

Finding an equation with parallel lines

You are given the following diagram:

You are also told that line segment AB runs parallel to the following line: y=0.5x5. Point A is at (2;3). Determine the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by recalling that the gradient of parallel lines is equal.


STEP: Determine the gradient of line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of the line AB. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line AB:

y=mx+cy=(0.5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(3)=(0.5)×(2)+cc=2

STEP: Write down the equation of the line AB in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=0.5x2

Submit your answer as:

ID is: 3107 Seed is: 70

Find the equation of a straight line

There are two points, P(34;4) and N(4;74). What is the equation of the straight line through the points P and N? Your answer should be exact (no rounding).

INSTRUCTION: Type the entire equation into the answer box.
Answer: The equation is: .
one-of
type(equation)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Think about the standard form of a straight line equation: is there anything we can substitute into the formula from the given information?
STEP: Substitute the values into the gradient formula
[−2 points ⇒ 3 / 5 points left]

The standard formula of a straight line is:

y=mx+c

where m is the slope of the graph and c is the point where the graph intercepts the y-axis. We are given only two points on the straight line. So we should use those points to try and find the slope and point where the graph intercepts the y-axis of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

m=y2y1x2x1

In this specific case we can represent each coordinate as follows:

x1=34x2=4y1=4y2=74

Now substitute these values into the formula to find the gradient.

gradient=(74)(4)4(34)=(234)(134)=(234)×(413)=(2313)

STEP: Now determine the c-value
[−2 points ⇒ 1 / 5 points left]

If we look at the standard form of a straight line equation again, we now have:

y=(2313)x+c
Where (2313) is the gradient of the graph. Also notice that this value is positive. This means that the straight line is increasing, moving from the bottom upwards. The next step is to solve for the c-value. In other words we need to find where the graph 'cuts' the y-axis. We can use any point on the straight line to substitute into our formula. Here we will use the given point: P(34;4).
y=(2313)x+c(4)=(2313)(34)+c(4)(6952)=cc=27752

The graph 'cuts' the y-axis at y=27752.


STEP: Now substitute all the calculated values into the standard formula
[−1 point ⇒ 0 / 5 points left]

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

y=2313x27752

Submit your answer as:

ID is: 3107 Seed is: 8099

Find the equation of a straight line

Compute the equation of the straight line through the points A(13;6) and B(6;4). Your answer should be exact (no rounding).

INSTRUCTION: Type the entire equation into the answer box.
Answer: The equation is: .
one-of
type(equation)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Think about the standard form of a straight line equation: is there anything we can substitute into the formula from the given information?
STEP: Substitute the values into the gradient formula
[−2 points ⇒ 3 / 5 points left]

The standard formula of a straight line is:

y=mx+c

where m is the gradient of the graph and c is the y-intercept. We are given only two points on the straight line. So we should use those points to try and find the slope and y-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

m=y2y1x2x1

In this specific case we can represent each coordinate as follows:

x1=13x2=6y1=6y2=4

Now substitute these values into the formula to find the gradient.

gradient=466(13)=(2)(193)=(2)×(319)=(619)

STEP: Now determine the c-value
[−2 points ⇒ 1 / 5 points left]

If we look at the standard form of a straight line equation again, we now have:

y=(619)x+c
Where (619) is the gradient of the graph. Also notice that this value is negative. This means that the straight line is decreasing, moving from the top downwards. The next step is to solve for the c-value. In other words we need to find where the graph 'cuts' the y-axis. We can use any point on the straight line to substitute into our formula. Here we will use the given point: A(13;6).
y=(619)x+c(6)=(619)(13)+c(6)(219)=cc=11219

The graph 'cuts' the y-axis at y=11219.


STEP: Now substitute all the calculated values into the standard formula
[−1 point ⇒ 0 / 5 points left]

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

y=619x+11219

Submit your answer as:

ID is: 3107 Seed is: 9568

Find the equation of a straight line

The points Q and R have the coordinates (5;43) and (34;43), respectively. Find the equation of the straight line through Q and R. Your answer should be exact (no rounding).

INSTRUCTION: Type the entire equation into the answer box.
Answer: The equation is: .
one-of
type(equation)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Think about the standard form of a straight line equation: is there anything we can substitute into the formula from the given information?
STEP: Substitute the values into the gradient formula
[−2 points ⇒ 3 / 5 points left]

The standard formula of a straight line is:

y=mx+c

where m is the slope of the graph and c is the y-intercept. We are given only two points on the straight line. So we should use those points to try and find the gradient and y-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

m=y2y1x2x1

In this specific case we can represent each coordinate as follows:

x1=5x2=34y1=43y2=43

Now substitute these values into the formula to find the gradient.

gradient=(43)(43)(34)(5)=0(234)=0

STEP: Now determine the c-value
[−2 points ⇒ 1 / 5 points left]

If we look at the standard form of a straight line equation again, we now have:

y=0x+c
Where 0 is the gradient of the graph. Also notice that this value is zero. This means that the straight line is a horizontal line, with a 'flat' gradient. The next step is to solve for the c-value. In other words we need to find where the graph 'cuts' the y-axis. We can use any point on the straight line to substitute into our formula. Here we will use the given point: Q(5;43).
y=0x+c(43)=0(5)+c(43)(0)=cc=43


STEP: Now substitute all the calculated values into the standard formula
[−1 point ⇒ 0 / 5 points left]

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

y=43

Submit your answer as:

3. Distance


ID is: 3114 Seed is: 7548

The distance formula with surds

Given the two points here, what is the distance between the points?

A(2;2)B(43;43)

Your answer should be exact. (This means the answer should either be a surd, like 'sqrt(15)', or a whole number. Do not round off your answer.)

Answer: d=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.


STEP: Substitute the values into the distance formula
[−1 point ⇒ 3 / 4 points left]

This is a distance formula question: we need to calculate the distance between the points (2;2) and (43;43). On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

x1=2;y1=2x2=43;y2=43

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

d=[x2x1]2+[y2y1]2=[(43)(2)]2+[(43)(2)]2

STEP: Simplify the square brackets and square them
[−2 points ⇒ 1 / 4 points left]

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative in the first square bracket which changes to addition.

d=[43+2]2+[432]2=[43+2][43+2]+[432][432]=(48+163+4)+(48163+4)=48+4+48+4

That simplified a lot! The key part is the cancellation of the surd terms, 163 and 163. From this point on there are only integer values to work with inside the root.


STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

d=52+52=104

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to 226, which is also accepted.)

The distance between the points is: 104.


Submit your answer as:

ID is: 3114 Seed is: 6207

The distance formula with surds

Given the two points here, what is the distance between the points?

M(2;210+1)N(2102;1)

Your answer should be exact. (This means the answer should either be a surd, like 'sqrt(15)', or a whole number. Do not round off your answer.)

Answer: d=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.


STEP: Substitute the values into the distance formula
[−1 point ⇒ 3 / 4 points left]

This is a distance formula question: we need to calculate the distance between the points (2;210+1) and (2102;1). On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

x1=2;y1=210+1x2=2102;y2=1

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

d=[x2x1]2+[y2y1]2=[(2102)(2)]2+[(1)(210+1)]2

STEP: Simplify the square brackets and square them
[−2 points ⇒ 1 / 4 points left]

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we simplify the square brackets.

In this case, we have to distribute the negative sign in the second square bracket into the binomial 210+1. One of the binomial terms is negative, so it becomes positive and the other term becomes negative. Additionally, there is a double negative in the first square bracket which simplifies to addition.

d=[2102+2]2+[1+2101]2=[210]2+[210]2=(210)(210)+(210)(210)=410+410

Brilliant: the trinomials in the square brackets each simplified into a single term and squaring the terms cancelled the roots. From this point on there are only integer values to work with inside the root.


STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

d=40+40=80

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to 45, which is also accepted.)

The distance between the points is: 80.


Submit your answer as:

ID is: 3114 Seed is: 7568

The distance formula with surds

What is the distance between these two points?

M(3;3)N(62;62)

Your answer should be exact. (This means the answer should either be a surd, like 'sqrt(15)', or a whole number. Do not round off your answer.)

Answer: d=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.


STEP: Substitute the values into the distance formula
[−1 point ⇒ 3 / 4 points left]

This is a distance formula question: we need to calculate the distance between the points (3;3) and (62;62). On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

x1=3;y1=3x2=62;y2=62

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

d=[x2x1]2+[y2y1]2=[(62)(3)]2+[(62)(3)]2

STEP: Simplify the square brackets and square them
[−2 points ⇒ 1 / 4 points left]

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative in the second square bracket which changes to addition.

d=[623]2+[62+3]2=[623][623]+[62+3][62+3]=(72+362+9)+(72362+9)=72+9+72+9

That simplified a lot! The key part is the cancellation of the surd terms, 362 and 362. From this point on there are only integer values to work with inside the root.


STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

d=81+81=162

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to 92, which is also accepted.)

The distance between the points is: 162.


Submit your answer as:

ID is: 3111 Seed is: 8778

Getting to know the distance formula

Consider two points: V(2;4) and W(2;4). Suppose that we want to find the distance between these points. The following shows the distance formula followed by the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2d=(22)2+((4)4)2
  1. In the distance formula, what does the term (y2y1) correspond to?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The expression (y2y1) is the difference in the y coordinates for the two points.


    STEP: Draw a sketch to see the details
    [−1 point ⇒ 0 / 1 points left]

    The question is about the distance formula. There are two coordinate pairs given, but this question asks about the distance formula itself. We need to decide the meaning of the term (y2y1) in the formula. This is easiest to see by graphing the points.

    The points are one above the other! This is because they have equal x coordinates, so they sit at the same x-position: the points are both on the line x=2, as shown in the diagram. The points are separated in the vertical direction, but they are not separated in the horizontal direction. Notice that this means the total distance between the points is exactly equal to the vertical distance between the points, as shown on the diagram.

    As the diagram shows, the (y2y1) term in the distance formula corresponds to the vertical distance between the points.

    The correct answer choice is: the vertical distance between the points.


    Submit your answer as:
  2. Complete the calculation to find the distance between the points given in Question 1.

    INSTRUCTION: Round your answer to two decimal points if appropriate.
    Answer: d=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You can use the distance formula if you want to, but it is not necessary in this case.


    STEP: Read the answer off the diagram (or calculate it)
    [−2 points ⇒ 0 / 2 points left]

    We can use the formula, but if the distance is vertical or horizontal, it is easy to just count the spaces between the points on the diagram above.

    distance=vertical distance=count from V to W=8

    The distance between points V and W is 8.

    We can of course use the formula to get the same answer. Start with the working shown in Question 1 and continue it.

    d=(x2x1)2+(y2y1)2=(22)2+((4)4)2=(0)2+(8)2

    The zero in this calculation comes from the fact that the points are not separated in the horizontal direction: the x coordinates of the points cancel. With that zero, the first term disappears because 02 is zero, which is exactly why we could ignore it above.

    d=(8)2=64=8

    Notice that the square root and the square effectively cancelled each other, although the square makes sure that the answer we get is positive (distances must be positive).

    The distance between the points is 8.


    Submit your answer as:

ID is: 3111 Seed is: 1760

Getting to know the distance formula

Consider two points: P(3;0) and Q(10;9). Suppose that we want to find the distance between these points. The following shows the distance formula followed by the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2d=(103)2+(90)2
  1. What is the distance formula based on?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think of squaring both sides of the distance formula to get this:

    d2=(x2x1)2+(y2y1)2

    What formula do you know which looks similar?


    STEP: Draw a sketch to see the details
    [−1 point ⇒ 0 / 1 points left]

    The question is about the distance formula. There are two coordinate pairs given, but we do not need them for this question. Instead, we need to identify the foundation of the distance formula (what it is based on). This is easiest to see by graphing the points.

    As you can see, we have built a right-angled triangle onto the points. We The distance between the points is the hypotenuse of the triangle. The quantity (x2x1) corresponds to the horizontal leg of the triangle and (y2y1) corresponds to the vertical leg of the triangle. Since this is a right-angled triangle, the theorem of Pythagoras applies to the lengths of the sides. In fact, the distance formula is built on the theorem of Pythagoras!

    The theorem of Pythagoras says that:

    c2=a2+b2

    where c is the hypotenuse of the triangle. The hypotenuse of the triangle above is the distance between the points. The other sides of the triangle correspond to sides a and b in Pythagoras's equation. Substituting the three sides of the triangle in the diagram above into the theorem, we get:

    c2=a2+b2distance2=(x2x1)2+(y2y1)2

    If we square root both sides, we get the distance formula:

    distance=(x2x1)2+(y2y1)2

    The correct answer choice is: the theorem of Pythagoras.


    Submit your answer as:
  2. Complete the calculation to find the distance between the points given in Question 1.

    INSTRUCTION: Round your answer to two decimal points if appropriate.
    Answer: d=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Continue the working started in the answer above.


    STEP: Work out the answer following BODMAS
    [−3 points ⇒ 0 / 3 points left]

    To determine the distance between the points, we need to complete the calculation which was started in Question 1. We already had:

    d=(x2x1)2+(y2y1)2=(103)2+(90)2

    Remember to follow BODMAS, and that a root acts like brackets (this means you must do all of the calculations under the root before you calculate the square root itself).

    d=(103)2+(90)2=(7)2+(9)2

    We still cannot evaluate the root: we must finish all calculations inside the root first, and then take the square root of that result.

    d=49+81=130=11.40175...11.4

    130 is not a perfect square, so we need to round the decimal to two places (according to the instructions).

    The distance between the points is 11.4.


    Submit your answer as:

ID is: 3111 Seed is: 9410

Getting to know the distance formula

Consider two points: V(2;2) and W(4;8). Suppose that we want to find the distance between these points. The following shows the distance formula followed by the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2d=(2(4))2+((2)(8))2
  1. What is the distance formula based on?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think of squaring both sides of the distance formula to get this:

    d2=(x2x1)2+(y2y1)2

    What formula do you know which looks similar?


    STEP: Draw a sketch to see the details
    [−1 point ⇒ 0 / 1 points left]

    The question is about the distance formula. There are two coordinate pairs given, but we do not need them for this question. Instead, we need to identify the foundation of the distance formula (what it is based on). This is easiest to see by graphing the points.

    As you can see, we have built a right-angled triangle onto the points. We The distance between the points is the hypotenuse of the triangle. The quantity (x2x1) corresponds to the horizontal leg of the triangle and (y2y1) corresponds to the vertical leg of the triangle. Since this is a right-angled triangle, the theorem of Pythagoras applies to the lengths of the sides. In fact, the distance formula is built on the theorem of Pythagoras!

    The theorem of Pythagoras says that:

    c2=a2+b2

    where c is the hypotenuse of the triangle. The hypotenuse of the triangle above is the distance between the points. The other sides of the triangle correspond to sides a and b in Pythagoras's equation. Substituting the three sides of the triangle in the diagram above into the theorem, we get:

    c2=a2+b2distance2=(x2x1)2+(y2y1)2

    If we square root both sides, we get the distance formula:

    distance=(x2x1)2+(y2y1)2

    The correct answer choice is: the theorem of Pythagoras.


    Submit your answer as:
  2. Complete the calculation to find the distance between the points given in Question 1.

    INSTRUCTION: Round your answer to two decimal points if appropriate.
    Answer: d=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Continue the working started in the answer above.


    STEP: Work out the answer following BODMAS
    [−3 points ⇒ 0 / 3 points left]

    To determine the distance between the points, we need to complete the calculation which was started in Question 1. We already had:

    d=(x2x1)2+(y2y1)2=(2(4))2+((2)(8))2

    Remember to follow BODMAS, and that a root acts like brackets (this means you must do all of the calculations under the root before you calculate the square root itself).

    d=(2(4))2+(2(8))2=(2+4)2+(2+8)2=(6)2+(6)2
    NOTE: In the first line the double negatives become addition

    We still cannot evaluate the root: we must finish all calculations inside the root first, and then take the square root of that result.

    d=36+36=72=8.48528...8.49

    72 is not a perfect square, so we need to round the decimal to two places (according to the instructions).

    The distance between the points is 8.49.


    Submit your answer as:

ID is: 1219 Seed is: 8849

Calculating a line length from two points

The following diagram shows two points on the Cartesian plane. Point A is at (−2;2) and point B is at (1;−4).

Calculate the length of line segment AB, correct to 2 decimal places.

Answer: Length of segment AB is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−1 point ⇒ 1 / 2 points left]

First we recall the equation for distance:

dAB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula and determine the distance
[−1 point ⇒ 0 / 2 points left]

Substituting into the equation, we get:

length =(1(2))2+(4(2))2=(1+2)2+(42)2=(3)2+(6)2=9+36=45=6.70820...

The length of the line segment, rounded to two decimal places, is 6.71.


Submit your answer as:

ID is: 1219 Seed is: 8641

Calculating a line length from two points

The following diagram shows two points on the Cartesian plane. Point A is at (−2;1) and point B is at (4;4).

Calculate the length of line segment AB, correct to 2 decimal places.

Answer: Length of segment AB is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−1 point ⇒ 1 / 2 points left]

First we recall the equation for distance:

dAB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula and determine the distance
[−1 point ⇒ 0 / 2 points left]

Substituting into the equation, we get:

length =(4(2))2+(4(1))2=(4+2)2+(41)2=(6)2+(3)2=36+9=45=6.70820...

The length of the line segment, rounded to two decimal places, is 6.71.


Submit your answer as:

ID is: 1219 Seed is: 1906

Calculating a line length from two points

The following diagram shows two points on the Cartesian plane. Point A is at (−1;3.5) and point B is at (3;−2.5).

Calculate the length of line segment AB, correct to 2 decimal places.

Answer: Length of segment AB is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−1 point ⇒ 1 / 2 points left]

First we recall the equation for distance:

dAB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula and determine the distance
[−1 point ⇒ 0 / 2 points left]

Substituting into the equation, we get:

length =(3(1))2+(2.5(3.5))2=(3+1)2+(2.53.5)2=(4)2+(6)2=16+36=52=7.21110...

The length of the line segment, rounded to two decimal places, is 7.21.


Submit your answer as:

ID is: 1206 Seed is: 5756

Finding a coordinate from a distance

The following picture shows two points on the Cartesian plane, A and B.

The line segment AB¯ has a length of 9.0139. Calculate the missing coordinate of point B. Round your answer to the nearest half-integer.

Answer: The coordinates of point B are: (2;).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−0 points ⇒ 3 / 3 points left]

First we recall the equation for distance:

AB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula
[−1 point ⇒ 2 / 3 points left]

Substituting into the equation, we get:

9.0139=(2(3))2+(y(4))29.0139=(2+3)2+(y4)2

STEP: Solve for y
[−1 point ⇒ 1 / 3 points left]

Now we re-arrange, and solve for the value of y:

(9.0139)2=(2+3)2+(y4)281.25=(2+3)2+(y4)281.25=(y4)2+25(y4)2=56.25y4=±56.25y=±7.5+4y=(12) or (3.5)

STEP: Interpret the result and write down the answer
[−1 point ⇒ 0 / 3 points left]

We now have a choice between 2 values for y. From the diagram we can see that the appropriate value for this question is 3.5.


Submit your answer as:

ID is: 1206 Seed is: 2879

Finding a coordinate from a distance

The following picture shows two points on the Cartesian plane, A and B.

The line segment AB¯ has a length of 6.7082. Calculate the missing coordinate of point B. Round your answer to the nearest half-integer.

Answer: The coordinates of point B are: (;2.5).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−0 points ⇒ 3 / 3 points left]

First we recall the equation for distance:

AB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula
[−1 point ⇒ 2 / 3 points left]

Substituting into the equation, we get:

6.7082=(x(1))2+(2.5(3.5))26.7082=(x+1)2+(2.53.5)2

STEP: Solve for x
[−1 point ⇒ 1 / 3 points left]

Now we re-arrange, and solve for the value of x:

(6.7082)2=(x+1)2+(2.53.5)245=(x+1)2+(2.53.5)245=(x+1)2+36(x+1)2=9x+1=±9x=±31x=(2) or (4)

STEP: Interpret the result and write down the answer
[−1 point ⇒ 0 / 3 points left]

We now have a choice between 2 values for x. From the diagram we can see that the appropriate value for this question is 2.


Submit your answer as:

ID is: 1206 Seed is: 5583

Finding a coordinate from a distance

The following picture shows two points on the Cartesian plane, A and B.

The distance between the points is 4.4721. Calculate the missing coordinate of point B. Round your answer to the nearest half-integer.

Answer: The coordinates of point B are: (;3.5).
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−0 points ⇒ 3 / 3 points left]

First we recall the equation for distance:

AB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula
[−1 point ⇒ 2 / 3 points left]

Substituting into the equation, we get:

4.4721=(x(1))2+(3.5(0.5))24.4721=(x+1)2+(3.5+0.5)2

STEP: Solve for x
[−1 point ⇒ 1 / 3 points left]

Now we re-arrange, and solve for the value of x:

(4.4721)2=(x+1)2+(3.5+0.5)220=(x+1)2+(3.5+0.5)220=(x+1)2+16(x+1)2=4x+1=±4x=±21x=(1) or (3)

STEP: Interpret the result and write down the answer
[−1 point ⇒ 0 / 3 points left]

We now have a choice between 2 values for x. From the diagram we can see that the appropriate value for this question is 1.


Submit your answer as:

ID is: 3112 Seed is: 9337

The distance formula: substituting correctly

Consider two points: C(9;3) and D(6;3). We want to find the distance between these points. The following shows the distance formula and the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2=(9(6))2+(3(3))2

In the distance formula, the coordinates (x1;y1) represent one of the points while (x2;y2) represents the other point. How do you know which of the given points, C or D, is (x1;y1)?

Answer: The correct choice is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the distance formula is about how far apart the points are. If you move your finger from point C to D, or from D to C, would the distance be different?


STEP: Consider the options and select the correct choice
[−1 point ⇒ 0 / 1 points left]

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and asks us this: how do we know which of the points (9;3) and (6;3) correspond to the points (x1;y1) and (x2;y2) in the formula? In other words, how do we know which of the points is the "starting" point?

Let's step back and look at what the question is about. The diagram below shows points C and D on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now getting back to the question at hand: how do we know which of the given points is (x1;y1)? Consider substituting the x-coordinates for the points, 9 and −6, into the term (x2x1): if you substitute ((6)9)2 you get (15)2, while if you substitute (9(6))2 you get (15)2. As soon as we calculate the squares, we will get the same (positive) answer either way! The same thing is true for the y calculation. So it does not matter which point you chose for (x1;y1) and (x2;y2) because of the squares in the distance formula will always give us positive values. Both of the substitutions here are correct.

d=((6)9)2+((3)3)2-- OR --d=(9(6))2+(3(3))2

Both of these calculations lead to the correct distance, which is 261.

The correct answer choice is: It does not matter which point you use for (x1;y1).


Submit your answer as:

ID is: 3112 Seed is: 5720

The distance formula: substituting correctly

Consider two points: C(5;4) and D(3;7). We want to find the distance between these points. According to one of your classmates, Adedapo, the following work is the beginning of the solution.

d=(x2x1)2+(y2y1)2=(3(5))2+(7(4))2

Are Adedapo's substitutions into the formula correct or not?

Answer: The correct choice is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the coordinates in the work: are they substituted in the correct places?


STEP: Consider the options and select the correct choice
[−1 point ⇒ 0 / 1 points left]

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points C and D on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? In this case, yes they are. In the distance formula, (x1;y1) represents one of the points, and (x2;y2) represents the other point. Adedapo is using point C as (x1;y1) and D as (x2;y2). He could do it the other way around, and that would be correct too. In other words, both of the following are correct:

d=(3(5))2+(7(4))2-- OR --d=((5)3)2+((4)7)2

Both of these calculations lead to the correct distance, which is 185.

The correct answer choice is: Yes, the substitutions are correct.


Submit your answer as:

ID is: 3112 Seed is: 7629

The distance formula: substituting correctly

Consider two points: C(7;6) and D(5;7). We want to find the distance between these points. According to one of your classmates, Akinbode, the following work is the beginning of the solution.

d=(x2x1)2+(y2y1)2=(7(5))2+((7)6)2

Are Akinbode's substitutions into the formula correct or not?

Answer: The correct choice is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the coordinates in the work: are they substituted in the correct places?


STEP: Consider the options and select the correct choice
[−1 point ⇒ 0 / 1 points left]

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points C and D on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? Shame, but Akinbode's substitutions are not correct because he mixed up the coordinates. In the distance formula, (x1;y1) represents one of the points, and (x2;y2) represents the other point. While it does not matter if you pick point C or D as (x1;y1), the substitutions must be consistent. Akinbode's substitutions are not consistent: the x-coordinates and the y-coordinates are substituted backwards compared to each other. Either the numbers 5 and 7 should be swapped or the numbers 6 and 7 should be swapped. The correct substitution can be either of the following:

d=((5)7)2+((7)6)2-- OR --d=(7(5))2+(6(7))2

Both of these calculations lead to the correct distance, which is 313.

You might notice that Akinbode's substitutions are a combination of the two correct versions shown above. Because of that, Akinbode would actually be lucky and get the correct answer. However, this is only because the squares in the formula cancel out his error. Even while he may be lucky to get the correct answer, the substitutions are not fully correct.

The correct answer choice is: No, there is an error.


Submit your answer as:

4. Midpoint


ID is: 1216 Seed is: 8476

Finding the mid-point of two points

The following diagram shows points A(1;1.5) and B(3;2.5). Point M is the mid-point of A and B.

Calculate the coordinates of the mid-point, M.

INSTRUCTION: Remember to type your answer with brackets and with ';' between the coordinates, for example: ( 3/2 ; -4).
Answer: The coordinates of M are: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Write down the mid-point formula. (If you do not remember it, look it up in your notes or in a book.)
STEP: Write down the mid-point formula
[−1 point ⇒ 2 / 3 points left]

Start with the mid-point formula:

M=(x1+x22;y1+y22)

In this formula, M is the mid-point of two other points. The coordinates in the formula, (x1;y1) and (x2;y2), are the coordinates of those other points.


STEP: Identify the values of x1,x2,y1 and y2 and substitute into the formula
[−1 point ⇒ 1 / 3 points left]

In this question, (x1;y1) and (x2;y2) must be the coordinates of points A and B. Therefore:

x1=1;y1=1.5;x2=3;andy2=2.5

Now substitute these values into the formula:

M=((1)+(3)2;(1.5)+(2.5)2)

STEP: Evaluate the answers
[−1 point ⇒ 0 / 3 points left]

Working out the calculations, we get the final answers:

M=((1)+(3)2;(1.5)+(2.5)2)=(22;12)=(1;12)

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point M agree with the coordinates (1;12)? Indeed, that looks quite reasonable!

The coordinates of the mid-point are (1;12).


Submit your answer as:

ID is: 1216 Seed is: 897

Finding the mid-point of two points

The following diagram shows points A(1;0) and B(1;4). Point M is the mid-point of A and B.

Calculate the coordinates of the mid-point, M.

INSTRUCTION: Remember to type your answer with brackets and with ';' between the coordinates, for example: ( 3/2 ; -4).
Answer: The coordinates of M are: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Write down the mid-point formula. (If you do not remember it, look it up in your notes or in a book.)
STEP: Write down the mid-point formula
[−1 point ⇒ 2 / 3 points left]

Start with the mid-point formula:

M=(x1+x22;y1+y22)

In this formula, M is the mid-point of two other points. The coordinates in the formula, (x1;y1) and (x2;y2), are the coordinates of those other points.


STEP: Identify the values of x1,x2,y1 and y2 and substitute into the formula
[−1 point ⇒ 1 / 3 points left]

In this question, (x1;y1) and (x2;y2) must be the coordinates of points A and B. Therefore:

x1=1;y1=0;x2=1;andy2=4

Now substitute these values into the formula:

M=((1)+(1)2;(0)+(4)2)

STEP: Evaluate the answers
[−1 point ⇒ 0 / 3 points left]

Working out the calculations, we get the final answers:

M=((1)+(1)2;(0)+(4)2)=(02;42)=(0;2)

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point M agree with the coordinates (0;2)? Indeed, that looks quite reasonable!

The coordinates of the mid-point are (0;2).


Submit your answer as:

ID is: 1216 Seed is: 812

Finding the mid-point of two points

The following diagram shows points A(1;2.5) and B(2;3.5). Point M is the mid-point of A and B.

Calculate the coordinates of the mid-point, M.

INSTRUCTION: Remember to type your answer with brackets and with ';' between the coordinates, for example: ( 3/2 ; -4).
Answer: The coordinates of M are: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Write down the mid-point formula. (If you do not remember it, look it up in your notes or in a book.)
STEP: Write down the mid-point formula
[−1 point ⇒ 2 / 3 points left]

Start with the mid-point formula:

M=(x1+x22;y1+y22)

In this formula, M is the mid-point of two other points. The coordinates in the formula, (x1;y1) and (x2;y2), are the coordinates of those other points.


STEP: Identify the values of x1,x2,y1 and y2 and substitute into the formula
[−1 point ⇒ 1 / 3 points left]

In this question, (x1;y1) and (x2;y2) must be the coordinates of points A and B. Therefore:

x1=1;y1=2.5;x2=2;andy2=3.5

Now substitute these values into the formula:

M=((1)+(2)2;(2.5)+(3.5)2)

STEP: Evaluate the answers
[−1 point ⇒ 0 / 3 points left]

Working out the calculations, we get the final answers:

M=((1)+(2)2;(2.5)+(3.5)2)=(12;12)

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point M agree with the coordinates (12;12)? Indeed, that looks quite reasonable!

The coordinates of the mid-point are (12;12).


Submit your answer as:

ID is: 3110 Seed is: 4793

Finding the mid-point from coordinates

Find the mid-point of the points Q(6;74) and R(6;1). Your answer should be exact (no rounding).

Your answer should be in brackets, with the ';' symbol between the coordinates. For example: ( 1/2 ; -5).

Answer: The mid-point is: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Use the mid-point formula. Start by figuring out the values of x1,x2,y1 and y2.
STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 3 points left]

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

M=(x1+x22;y1+y22)

In this equation, M represents the mid-point, which has the coordinates (xM;yM), while the variables x1, etc. refer to the coordinates of the end points. In this case, the end points are the points Q and R. We must decide which point to use as point 1 and which to use as point 2. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

point 1:(6;74)point 2:(6;1)

This means:

x1=6x2=6y1=74y2=1

Plugging these values into the mid-point formula, we get:

(xM;yM)=((6)+62;(74)+(1)2)

STEP: Substitute and evaluate each expression
[−2 points ⇒ 0 / 3 points left]

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the x coordinates and the other for the y coordinates. It is helpful to separate the calculations, like this:

x coordinatesy coordinatesxM=(6)+62yM=(74)+(1)2=02=(34)2=0=(34)×12=38

The resulting coordinates for the mid-point are (0;38). Notice that the x coordinate of the mid-point is zero, which happens because the x coordinates of Q and R are opposite values.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points Q and R.

The mid-point of Q and R is at: (0;38).


Submit your answer as:

ID is: 3110 Seed is: 6404

Finding the mid-point from coordinates

Compute the mid-point of the points Q(143;83) and R(2;143). Your answer should be exact (no rounding).

Your answer should be in brackets, with the ';' symbol between the coordinates. For example: ( 1/2 ; -5).

Answer: The mid-point is: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Use the mid-point formula. Start by figuring out the values of x1,x2,y1 and y2.
STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 3 points left]

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

M=(x1+x22;y1+y22)

In this equation, M represents the mid-point, which has the coordinates (xM;yM), while the variables x1, etc. refer to the coordinates of the end points. In this case, the end points are the points Q and R. We must decide which point to use as point 1 and which to use as point 2. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

point 1:(143;83)point 2:(2;143)

This means:

x1=143x2=2y1=83y2=143

Plugging these values into the mid-point formula, we get:

(xM;yM)=((143)+22;(83)+(143)2)

STEP: Substitute and evaluate each expression
[−2 points ⇒ 0 / 3 points left]

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the x coordinates and the other for the y coordinates. It is helpful to separate the calculations, like this:

x coordinatesy coordinatesxM=(143)+22yM=(83)+(143)2=(83)2=22=(83)×12=1=43

The resulting coordinates for the mid-point are (43;1).

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points Q and R.

The mid-point of Q and R is at: (43;1).


Submit your answer as:

ID is: 3110 Seed is: 9913

Finding the mid-point from coordinates

The points Q and R have the coordinates (163;1) and (1;72), respectively. Determine the coordinates of the mid-point of Q and R. Your answer should be exact (no rounding).

Your answer should be in brackets, with the ';' symbol between the coordinates. For example: ( 1/2 ; -5).

Answer: The mid-point is: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Use the mid-point formula. Start by figuring out the values of x1,x2,y1 and y2.
STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 3 points left]

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

M=(x1+x22;y1+y22)

In this equation, M represents the mid-point, which has the coordinates (xM;yM), while the variables x1, etc. refer to the coordinates of the end points. In this case, the end points are the points Q and R. We must decide which point to use as point 1 and which to use as point 2. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

point 1:(163;1)point 2:(1;72)

This means:

x1=163x2=1y1=1y2=72

Plugging these values into the mid-point formula, we get:

(xM;yM)=((163)+(1)2;(1)+(72)2)

STEP: Substitute and evaluate each expression
[−2 points ⇒ 0 / 3 points left]

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the x coordinates and the other for the y coordinates. It is helpful to separate the calculations, like this:

x coordinatesy coordinatesxM=(163)+(1)2yM=(1)+(72)2=(133)2=(52)2=(133)×12=(52)×12=136=54

The resulting coordinates for the mid-point are (136;54).

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points Q and R.

The mid-point of Q and R is at: (136;54).


Submit your answer as:

ID is: 3109 Seed is: 176

Working with algebraic coordinates and the mid-point

The mid-point of a Point N(a;2b+2) and another Point P is at (2a;2b+32). What are the coordinates of Point P?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of P is .
  2. The y-coordinate of P is .
expression
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.


STEP: Organise the information and make a plan
[−1 point ⇒ 3 / 4 points left]

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point P.

Point P:(xP;yP)Point N:(a;2b+2)the mid-point:(2a;2b+32)

We do not know the coordinates of Point P, so we use variables to represent them.

The coordinates are given in terms of a and b, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

M=(x1+x22;y1+y22)

STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

(xM;yM)=(x1+x22;y1+y22)(2a;2b+32)=(xP+a2;yP+(2b+2)2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The result above represents two separate equations: one for x and another for y. We can write them separately and solve each one. Begin by multiplying the equations by 2 to cancel the denominators.

x-coordinates2a=xP+a22(2a)=xP+a4a=xP+a4aa=xP3a=xPy-coordinates2b+32=yP+(2b+2)22(2b+32)=yP+2b+24b+3=yP+2b+24b+32b2=yP2b+1=yP

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point P: (3a;2b+1).

The final answers are:

  1. The x-coordinate of P is 3a.
  2. The y-coordinate of P is 2b+1.

Submit your answer as: and

ID is: 3109 Seed is: 7357

Working with algebraic coordinates and the mid-point

What are the coordinates of a point N if the mid-point of N and another point P is at (x;3) and the coordinates of P are (x4;3y+3)?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of N is .
  2. The y-coordinate of N is .
expression
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.


STEP: Organise the information and make a plan
[−1 point ⇒ 3 / 4 points left]

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point N.

Point P:(x4;3y+3)Point N:(xN;yN)the mid-point:(x;3)

We do not know the coordinates of Point N, so we use variables to represent them.

The coordinates are given in terms of x and y, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

M=(x1+x22;y1+y22)

STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

(xM;yM)=(x1+x22;y1+y22)(x;3)=((x4)+xN2;(3y+3)+yN2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The result above represents two separate equations: one for x and another for y. We can write them separately and solve each one. Begin by multiplying the equations by 2 to cancel the denominators.

x-coordinatesx=(x4)+xN22(x)=x4+xN2x=x4+xN2x+x+4=xN3x+4=xNy-coordinates3=(3y+3)+yN22(3)=3y+3+yN6=3y+3+yN6+3y3=yN3y+3=yN

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point N: (3x+4;3y+3).

The final answers are:

  1. The x-coordinate of N is 3x+4.
  2. The y-coordinate of N is 3y+3.

Submit your answer as: and

ID is: 3109 Seed is: 1681

Working with algebraic coordinates and the mid-point

What are the coordinates of a point P if the mid-point of P and another point N is at (2x21;12) and the coordinates of N are (2x21;y2+1)?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of P is .
  2. The y-coordinate of P is .
expression
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.


STEP: Organise the information and make a plan
[−1 point ⇒ 3 / 4 points left]

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point P.

Point P:(xP;yP)Point N:(2x21;y2+1)the mid-point:(2x21;12)

We do not know the coordinates of Point P, so we use variables to represent them.

The coordinates are given in terms of x and y, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

M=(x1+x22;y1+y22)

STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

(xM;yM)=(x1+x22;y1+y22)(2x21;12)=(xP+(2x21)2;yP+(y2+1)2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The result above represents two separate equations: one for x and another for y. We can write them separately and solve each one. Begin by multiplying the equations by 2 to cancel the denominators.

x-coordinates2x21=xP+(2x21)22(2x21)=xP+2x214x22=xP+2x214x222x2+1=xP2x21=xPy-coordinates12=yP+(y2+1)22(12)=yP+y2+11=yP+y2+11y21=yPy2=yP

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point P: (2x21;y2).

The final answers are:

  1. The x-coordinate of P is 2x21.
  2. The y-coordinate of P is y2.

Submit your answer as: and

ID is: 3108 Seed is: 5642

Using a mid-point to find an end point

As shown on the picture below, the mid-point of B and another Point A is at (14;12) and the coordinates of A are (32;6).

What are the coordinates of a Point B?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of B is .
  2. The y-coordinate of B is .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the question includes a mid-point, think about how you can use the mid-point formula.


STEP: Set up the equations based on the mid-point formula
[−2 points ⇒ 2 / 4 points left]

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

Point A:(32;6)Point B:(xB;yB)the mid-point:(14;12)

We do not know the coordinates of Point B, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

M=(x1+x22;y1+y22)(xM;yM)=(x1+x22;y1+y22)

In this case, we know that the mid-point coordinates are (14;12), but we do not know the coordinates of Point B. So the equation becomes:

(14;12)=((32)+xB2;(6)+yB2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The equation above actually represents two separate equations: one for x and another for y. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. The x equation has an LCD of 4 while the y equation will be multiplied by 2. (Notice that when the 4 cancels the denominator of 2, it leaves a factor of 2 which we must distribute into the binomial.)

x-coordinates14=(32)+xB24(14)=((32)+xB2)41=(32+xB)21=3+2xB1+3=2xB4=2xB2=xBy-coordinates12=(6)+yB22(12)=((6)+yB2)21=6+yB1+6=yB7=yB

We get the answers xB = 2 and yB = 7. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point B on the graph.

Point B must be at (2;7), which means the final answers are:

  1. The x-coordinate of B is 2.
  2. The y-coordinate of B is 7.

Submit your answer as: and

ID is: 3108 Seed is: 4243

Using a mid-point to find an end point

The mid-point of Point A(112;5) and another Point B is at (74;6). This is represented on the diagram below.

What are the coordinates of Point B?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of B is .
  2. The y-coordinate of B is .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the question includes a mid-point, think about how you can use the mid-point formula.


STEP: Set up the equations based on the mid-point formula
[−2 points ⇒ 2 / 4 points left]

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

Point A:(112;5)Point B:(xB;yB)the mid-point:(74;6)

We do not know the coordinates of Point B, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

M=(x1+x22;y1+y22)(xM;yM)=(x1+x22;y1+y22)

In this case, we know that the mid-point coordinates are (74;6), but we do not know the coordinates of Point B. So the equation becomes:

(74;6)=((112)+xB2;5+yB2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The equation above actually represents two separate equations: one for x and another for y. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. The x equation has an LCD of 4 while the y equation will be multiplied by 2. (Notice that when the 4 cancels the denominator of 2, it leaves a factor of 2 which we must distribute into the binomial.)

x-coordinates74=(112)+xB24(74)=((112)+xB2)47=(112+xB)27=11+2xB7+11=2xB4=2xB2=xBy-coordinates6=5+yB22(6)=(5+yB2)212=5+yB125=yB7=yB

We get the answers xB = 2 and yB = 7. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point B on the graph.

Point B must be at (2;7), which means the final answers are:

  1. The x-coordinate of B is 2.
  2. The y-coordinate of B is 7.

Submit your answer as: and

ID is: 3108 Seed is: 5211

Using a mid-point to find an end point

Consider two Points B and A. As shown on the diagram below, B is at (32;2) and the mid-point of B and A is (14;52).

Determine the coordinates of A.

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of A is .
  2. The y-coordinate of A is .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the question includes a mid-point, think about how you can use the mid-point formula.


STEP: Set up the equations based on the mid-point formula
[−2 points ⇒ 2 / 4 points left]

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

Point A:(xA;yA)Point B:(32;2)the mid-point:(14;52)

We do not know the coordinates of Point A, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

M=(x1+x22;y1+y22)(xM;yM)=(x1+x22;y1+y22)

In this case, we know that the mid-point coordinates are (14;52), but we do not know the coordinates of Point A. So the equation becomes:

(14;52)=(xA+(32)2;yA+(2)2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The equation above actually represents two separate equations: one for x and another for y. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. The x equation has an LCD of 4 while the y equation will be multiplied by 2. (Notice that when the 4 cancels the denominator of 2, it leaves a factor of 2 which we must distribute into the binomial.)

x-coordinates14=xA+(32)24(14)=(xA+(32)2)41=(xA32)21=2xA31+3=2xA4=2xA2=xAy-coordinates52=yA+(2)22(52)=(yA+(2)2)25=yA25+2=yA7=yA

We get the answers xA = 2 and yA = 7. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point A on the graph.

Point A must be at (2;7), which means the final answers are:

  1. The x-coordinate of A is 2.
  2. The y-coordinate of A is 7.

Submit your answer as: and

ID is: 3113 Seed is: 28

Mid-point calculations with algebra

The points A and B have the following coordinates:

A(3a;3b4)B(a;3b+4),a,bN

Determine the coordinates of the mid-point of A and B. Type your answer with brackets and the ';' symbol between the coordinates. For example, your answer might look like this: ( a + 3 ; -2b ).

Answer:
The mid-point coordinates are: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Use the mid-point formula. If the variables seem to make the question complex, try to ignore them and simply think about working with the formula.
STEP: Organise the information in the question
[−1 point ⇒ 3 / 4 points left]

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "a,bN" does not matter: for the mid-point it is the coordinate values which we need.)

point A:(3a;3b4)point B:(a;3b+4)the mid-point:(xM;yM)coordinates so we must use variablesWe do not know the mid-point

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.


STEP: Substitute into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points A and B. The mid-point formula is:

M=(x1+x22;y1+y22)

where x1,y1, etc. represent the coordinates of the end points (A and B). So:

x1=3a,x2=a,y1=3b4,y2=3b+4

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

(xM;yM)=((3a)+(a)2;(3b4)+(3b+4)2)

STEP: Simplify each expression
[−2 points ⇒ 0 / 4 points left]

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for x and one for y. We will separate these calculations to simplify things.

x coordinatesy coordinatesxM=(3a)+(a)2yM=(3b4)+(3b+4)2=3a+a2=3b4+3b+42=4a2=02=2a=0

There we have the answer! The mid-point of the two points is: (2a;0).


Submit your answer as:

ID is: 3113 Seed is: 8722

Mid-point calculations with algebra

Determine the mid-point of the points here:

Q(3a4;b24)R(3a4;b25),a,bN

Type your answer with brackets and the ';' symbol between the coordinates. For example, your answer might look like this: ( a + 3 ; -2b ).

Answer:
The mid-point coordinates are: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Use the mid-point formula. If the variables seem to make the question complex, try to ignore them and simply think about working with the formula.
STEP: Organise the information in the question
[−1 point ⇒ 3 / 4 points left]

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "a,bN" does not matter: for the mid-point it is the coordinate values which we need.)

point Q:(3a4;b24)point R:(3a4;b25)the mid-point:(xM;yM)coordinates so we must use variablesWe do not know the mid-point

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.


STEP: Substitute into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points Q and R. The mid-point formula is:

M=(x1+x22;y1+y22)

where x1,y1, etc. represent the coordinates of the end points (Q and R). So:

x1=3a4,x2=3a4,y1=b24,y2=b25

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

(xM;yM)=((3a4)+(3a4)2;(b24)+(b25)2)

STEP: Simplify each expression
[−2 points ⇒ 0 / 4 points left]

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for x and one for y. We will separate these calculations to simplify things.

x coordinatesy coordinatesxM=(3a4)+(3a4)2yM=(b24)+(b25)2=3a4+3a42=b24+b252=6a82=92=2(3a4)2=92=3a4

There we have the answer! The mid-point of the two points is: (3a4;92).


Submit your answer as:

ID is: 3113 Seed is: 4591

Mid-point calculations with algebra

The points P and N have the following coordinates:

P(3p;4q1)N(3p;2q+4),p,qZ

Determine the coordinates of the mid-point of P and N. Type your answer with brackets and the ';' symbol between the coordinates. For example, your answer might look like this: ( a + 3 ; -2b ).

Answer:
The mid-point coordinates are: .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Use the mid-point formula. If the variables seem to make the question complex, try to ignore them and simply think about working with the formula.
STEP: Organise the information in the question
[−1 point ⇒ 3 / 4 points left]

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "p,qZ" does not matter: for the mid-point it is the coordinate values which we need.)

point P:(3p;4q1)point N:(3p;2q+4)the mid-point:(xM;yM)coordinates so we must use variablesWe do not know the mid-point

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.


STEP: Substitute into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points P and N. The mid-point formula is:

M=(x1+x22;y1+y22)

where x1,y1, etc. represent the coordinates of the end points (P and N). So:

x1=3p,x2=3p,y1=4q1,y2=2q+4

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

(xM;yM)=((3p)+(3p)2;(4q1)+(2q+4)2)

STEP: Simplify each expression
[−2 points ⇒ 0 / 4 points left]

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for x and one for y. We will separate these calculations to simplify things.

x coordinatesy coordinatesxM=(3p)+(3p)2yM=(4q1)+(2q+4)2=3p3p2=4q12q+42=6p2=6q+32=3p=6q2+32=3q+32

There we have the answer! The mid-point of the two points is: (3p;3q+32).


Submit your answer as:

5. Angles


ID is: 1584 Seed is: 7819

Finding the equation of a specific line.

Determine the equation of a line passing through the point (4;6) which is perpendicular to the line

y=4x5+2

INSTRUCTION: Type the entire equation in standard form, which will either be y=mx+c or x=c.
Answer: The equation is: .
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: m=45.

For a line which is perpendicular to this, the gradient must be the negative reciprocal: m=54.


STEP: Find the y-intercept of the line
[−1 point ⇒ 1 / 3 points left]

Now substitute the coordinates (4;6), together with the gradient, into the equation to find the value of the y-intercept, c.

y=mx+c(6)=(54)(4)+c6=5+c11=c

STEP: Write the final equation
[−1 point ⇒ 0 / 3 points left]

Now we just put all of the pieces together into standard form. The equation that we want is

y=5x411

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (4;6) is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the lines are perpendicular to each other.

The equation of the line is y=5x411.


Submit your answer as:

ID is: 1584 Seed is: 7205

Finding the equation of a specific line.

Given the equation

x=4

what is the equation of another line which passes through the point (1;6) and is perpendicular to the given line?

INSTRUCTION: Type the entire equation in standard form, which will either be y=mx+c or x=c.
Answer: The equation is: .
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line we want. This is a special situation, because the equation given represents a vertical line (equations like 'x= a number' are always vertical lines). The gradient of a vertical line is undefined.

For a line which is perpendicular to this, we will get a horizontal line, which must have a gradient of zero.


STEP: Find the y-intercept of the line
[−1 point ⇒ 1 / 3 points left]

Now substitute the coordinates (1;6), together with the gradient, into the equation to find the value of the y-intercept, c.

y=mx+c(6)=(0)(1)+c6=0+c6=c

STEP: Write the final equation
[−1 point ⇒ 0 / 3 points left]

Now we just put all of the pieces together into standard form. The equation that we want is

y=6

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (1;6) is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that while the original equation (in blue) is vertical, the perpendicular line is horizontal.

The equation of the line is y=6.


Submit your answer as:

ID is: 1584 Seed is: 2723

Finding the equation of a specific line.

Given the equation

y=2x+4

what is the equation of another line which passes through the point (2;2) and is parallel to the given line?

INSTRUCTION: Type the entire equation in standard form, which will either be y=mx+c or x=c.
Answer: The equation is: .
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: m=2.

For a line which is parallel to this, we must use the same gradient: the gradient of the line we want is also m=2.


STEP: Find the y-intercept of the line
[−1 point ⇒ 1 / 3 points left]

Now substitute the coordinates (2;2), together with the gradient, into the equation to find the value of the y-intercept, c.

y=mx+c(2)=(2)(2)+c2=4+c2=c

STEP: Write the final equation
[−1 point ⇒ 0 / 3 points left]

Now we just put all of the pieces together into standard form. The equation that we want is

y=2x2

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (2;2) is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the two lines are parallel.

The equation of the line is y=2x2.


Submit your answer as:

ID is: 1212 Seed is: 2655

Finding equations of perpendicular lines

The graph here shows a line segment, AB¯. The blue dashed line is perpendicular to AB¯.

The equation of the blue dashed line is y=0.5x1.0. Point A is at (1;1).

Determine the equation of the line segment AB¯.

INSTRUCTION: Type only the right hand side of the equation into the box below.
Answer: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The gradient of the line AB is equal to: 1gradientPL, where PL stands for Perpendicular Line.


STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of AB¯. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of AB¯ must be equal to: 1gradientPL, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is 0.5.

y=mx+cy=(1gradientPL)x+cy=(10.5)x+cy=(2)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(1)=(2)(1)+cc=1

STEP: Write the equation of the line in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=2x1

Submit your answer as:

ID is: 1212 Seed is: 9943

Finding equations of perpendicular lines

The graph here shows a line segment, AB¯. The blue dashed line is perpendicular to AB¯.

The equation of the blue dashed line is y=0.5x+0.5. Point A is at (1;3.5).

Determine the equation of the line segment AB¯.

INSTRUCTION: Type only the right hand side of the equation into the box below.
Answer: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The gradient of the line AB is equal to: 1gradientPL, where PL stands for Perpendicular Line.


STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of AB¯. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of AB¯ must be equal to: 1gradientPL, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is 0.5.

y=mx+cy=(1gradientPL)x+cy=(10.5)x+cy=(2)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(3.5)=(2)(1)+cc=1.5

STEP: Write the equation of the line in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=2x+1.5

Submit your answer as:

ID is: 1212 Seed is: 7661

Finding equations of perpendicular lines

The graph here shows a line segment, AB¯. The blue dashed line is perpendicular to AB¯.

The equation of the blue dashed line is y=x+1.5. Point A is at (1;3).

Determine the equation of the line segment AB¯.

INSTRUCTION: Type only the right hand side of the equation into the box below.
Answer: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The gradient of the line AB is equal to: 1gradientPL, where PL stands for Perpendicular Line.


STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of AB¯. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of AB¯ must be equal to: 1gradientPL, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is 1.

y=mx+cy=(1gradientPL)x+cy=(11)x+cy=(1)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(3)=(1)(1)+cc=2

STEP: Write the equation of the line in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=x+2

Submit your answer as:

ID is: 2724 Seed is: 6173

Calculating the inclination of a line from two points

Calculate the inclination of the line (θ). Round your answer to 2 decimal places.

Answer: θ = °
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

To calculate the inclination of a straight line, we first need to calculate the gradient:

m=yByAxBxA=(3.5)(3.5)(5)(2)m=1

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now that we know the gradient, we use the following equation to determine the value of θ:

m=tan(θ)θ=tan1(m)=tan1(1)=45°

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.


STEP: <no title>
[−0 points ⇒ 0 / 2 points left]

If the gradient of the line is negative, we add 180° to the above angle to get an obtuse angle. In this case, the gradient is positive.

θ=45°

Submit your answer as:

ID is: 2724 Seed is: 1217

Calculating the inclination of a line from two points

Find the inclination of the line (θ). Round your answer to 2 decimal places.

Answer: θ = °
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the inclination of a straight line, we first need to calculate the gradient:

m=yByAxBxA=(0.5)(3.5)(1)(1)m=2

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now that we know the gradient, we use the following equation to determine the value of θ:

m=tan(θ)θ=tan1(m)=tan1(2)=63.43°

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

If the gradient of the line is negative, we add 180° to the above angle to get an obtuse angle. In this case, the gradient is negative. (You can see in the diagram that the angle of inclination is greater than 90 degrees.)

θ=63.43°+180°θ=116.57°

Submit your answer as:

ID is: 2724 Seed is: 7939

Calculating the inclination of a line from two points

Find the inclination of the line (θ). Round your answer to 2 decimal places.

Answer: θ = °
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

To calculate the inclination of a straight line, we first need to calculate the gradient:

m=yByAxBxA=(4)(1)(5)(5)m=0.5

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now that we know the gradient, we use the following equation to determine the value of θ:

m=tan(θ)θ=tan1(m)=tan1(0.5)=26.57°

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.


STEP: <no title>
[−0 points ⇒ 0 / 2 points left]

If the gradient of the line is negative, we add 180° to the above angle to get an obtuse angle. In this case, the gradient is positive.

θ=26.57°

Submit your answer as:

ID is: 1585 Seed is: 1009

Parallel and perpendicular lines

Are the following lines parallel, perpendicular, or neither?

y=x4y+33y3=5x15

Answer choices:

  • parallel
  • perpendicular
  • neither
INSTRUCTION: Type your answer from the three choices given; mispelled answers will be marked wrong.
Answer: The lines are .
string
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

y=mx+c

STEP: Arrange the first equation in standard form
[−1 point ⇒ 2 / 3 points left]

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

y=x4y+33y=x+313(3y)=(x+3)13y=x3+1

The gradient of the first equation is m1=13.


STEP: Arrange the second equation in standard form
[−1 point ⇒ 1 / 3 points left]

Now change the second equation into standard form.

3y3=5x153y=5x1213(3y)=(5x12)13y=5x34

The gradient of the second equation is m2=53.


STEP: Compare the gradients of the equations
[−1 point ⇒ 0 / 3 points left]

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

  • if the gradients are equal the lines are parallel
  • if the gradients are negative reciprocals the lines are perpendicular

For the equations in this question, the gradients are not equal, so the lines are not parallel. Also, the gradients are not negative reciprocals of each other, so the lines are not perpendicular.

The two equations are graphed on the Cartesian plane below. We can see that the lines are neither parallel nor perpendicular.

The correct answer is neither.


Submit your answer as:

ID is: 1585 Seed is: 1877

Parallel and perpendicular lines

Are the following lines parallel, perpendicular, or neither?

15x=6y+106=5x2y

Answer choices:

  • parallel
  • perpendicular
  • neither
INSTRUCTION: Type your answer from the three choices given; mispelled answers will be marked wrong.
Answer: The lines are .
string
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

y=mx+c

STEP: Arrange the first equation in standard form
[−1 point ⇒ 2 / 3 points left]

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

15x=6y+106y=15x+1016(6y)=(15x+10)16y=5x2+53

The gradient of the first equation is m1=52.


STEP: Arrange the second equation in standard form
[−1 point ⇒ 1 / 3 points left]

Now change the second equation into standard form.

6=5x2y2y=5x612(2y)=(5x6)12y=5x23

The gradient of the second equation is m2=52.


STEP: Compare the gradients of the equations
[−1 point ⇒ 0 / 3 points left]

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

  • if the gradients are equal the lines are parallel
  • if the gradients are negative reciprocals the lines are perpendicular

The equations have equal gradients, so they must be parallel lines.

The two equations are graphed on the Cartesian plane below. We can see that the two lines run parallel to one another.

The correct answer is parallel.


Submit your answer as:

ID is: 1585 Seed is: 5824

Parallel and perpendicular lines

Determine whether the equations here represent lines which are parallel, perpendicular, or neither.

2y=5x4y22x=5y

Answer choices:

  • parallel
  • perpendicular
  • neither
INSTRUCTION: Type your answer from the three choices given; mispelled answers will be marked wrong.
Answer: The lines are .
string
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

y=mx+c

STEP: Arrange the first equation in standard form
[−1 point ⇒ 2 / 3 points left]

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

2y=5x4y22y=5x212(2y)=(5x2)12y=5x21

The gradient of the first equation is m1=52.


STEP: Arrange the second equation in standard form
[−1 point ⇒ 1 / 3 points left]

Now change the second equation into standard form.

2x=5y5y=2x15(5y)=(2x)15y=2x5

The gradient of the second equation is m2=25.


STEP: Compare the gradients of the equations
[−1 point ⇒ 0 / 3 points left]

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

  • if the gradients are equal the lines are parallel
  • if the gradients are negative reciprocals the lines are perpendicular

The two gradients are negative reciprocals of each other. So the lines are perpendicular. (Notice also that the gradients have a product of 1: (52)(25) is equal to 1, which is always true for perpendicular gradients.)

The two equations are graphed on the Cartesian plane below. We can see that the two lines sit at right angles to each other: they are perpendicular.

The correct answer is perpendicular.


Submit your answer as:

ID is: 2692 Seed is: 7823

Calculating equations of perpendicular lines

Consider the following diagram:

Line AB is perpendicular to the line y=x+1,5 (blue line).

Find the equation of AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of a straight line equation is y=mx+c.

To calculate the equation of the straight line, we first need the gradient (m) of AB, which we can get from the perpendicular line:

For perpendicular lines, we know that m1×m2=1:

mAB×1=1mAB=11=1

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we substitute the coordinates of A into the equation to calculate the value of the y-intercept (c):

y=1x+c0=(1)(1)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is:

y=x+1

Submit your answer as:

ID is: 2692 Seed is: 8167

Calculating equations of perpendicular lines

Consider the following diagram:

Line AB is perpendicular to the line y=0,5x1,0 (blue line).

Calculate the equation of AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of a straight line equation is y=mx+c.

To calculate the equation of the straight line, we first need the gradient (m) of AB, which we can get from the perpendicular line:

For perpendicular lines, we know that m1×m2=1:

mAB×0.5=1mAB=10.5=2

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we substitute the coordinates of A into the equation to calculate the value of the y-intercept (c):

y=2x+c3=(2)(1)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is:

y=2x1

Submit your answer as:

ID is: 2692 Seed is: 8854

Calculating equations of perpendicular lines

Consider the following diagram:

Line AB is perpendicular to the line y=2x0,75 (blue line).

Calculate the equation of AB.

Answer: y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of a straight line equation is y=mx+c.

To calculate the equation of the straight line, we first need the gradient (m) of AB, which we can get from the perpendicular line:

For perpendicular lines, we know that m1×m2=1:

mAB×2=1mAB=12=0.5

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we substitute the coordinates of A into the equation to calculate the value of the y-intercept (c):

y=0.5x+c1=(0.5)(3)+cc=0.5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is:

y=0.5x0.5

Submit your answer as:

ID is: 2751 Seed is: 7760

Using inclination to find the equation of a line

The line shown below has an angle of inclination θ = 45°.

Find the equation of the line. Round your answer to one decimal place.

Answer:y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

We are given the angle of inclination of the line, which we can use to calculate the gradient:

m=tanθ=tan(45°)=1

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

We know the general form of the equation of a straight line is y=mx+c.

Substitute the value for m into the equation:

y=1x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

Next we calculate the value of c by substituting the coordinates of point A(2;3) into the equation and simplifying:

y=1x+c3=(1)(2)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Therefore, the equation of the line AB is

y=x1

Submit your answer as:

ID is: 2751 Seed is: 5633

Using inclination to find the equation of a line

The line shown below has an angle of inclination θ = 26.57°.

Determine the equation of the line. Round your answer to one decimal place.

Answer:y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

We are given the angle of inclination of the line, which we can use to calculate the gradient:

m=tanθ=tan(26.57°)=0.5

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

We know the general form of the equation of a straight line is y=mx+c.

Substitute the value for m into the equation:

y=0.5x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

Next we calculate the value of c by substituting the coordinates of point A(2;1.5) into the equation and simplifying:

y=0.5x+c1.5=(0.5)(2)+cc=0.5

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Therefore, the equation of the line AB is

y=0.5x0.5

Submit your answer as:

ID is: 2751 Seed is: 3983

Using inclination to find the equation of a line

The line shown below has an angle of inclination θ = 63.43°.

Find the equation of the line. Round your answer to one decimal place.

Answer:y =
expression
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

We are given the angle of inclination of the line, which we can use to calculate the gradient:

m=tanθ=tan(63.43°)=2

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

We know the general form of the equation of a straight line is y=mx+c.

Substitute the value for m into the equation:

y=2x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

Next we calculate the value of c by substituting the coordinates of point A(1;1.5) into the equation and simplifying:

y=2x+c1.5=(2)(1)+cc=0.5

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Therefore, the equation of the line AB is

y=2x+0.5

Submit your answer as:

ID is: 2776 Seed is: 3466

Finding the equation of a line

What is the equation of the line which is perpendicular to y=2x3+4 and passes through (2;3)?

Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.

Answer:
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. From the given equation you can see that the gradient is m=23.

For lines that are perpendicular, the gradient must be the negative reciprocal, therefore m=32.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Substitute the coordinates (2;3) and the gradient into the equation to find the value of c.

y=mx+c3=(32)(2)+c3=3+c6=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=3x26.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (2;3) is shown with a black dot labelled with an A. You can see that the red line passes through the point and that the lines are perpendicular to each other.


Submit your answer as:

ID is: 2776 Seed is: 1786

Finding the equation of a line

What is the equation of the line which is parallel to y=3x+2 and passes through (2;4)?

Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.

Answer:
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. From the given equation you can see that the gradient is m=3.

For lines that are parallel, we know that the gradients are equal so m=3.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Substitute the coordinates (2;4) and the gradient into the equation to find the value of c.

y=mx+c4=(3)(2)+c4=6+c10=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=3x10.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (2;4) is shown with a black dot labelled with an A. You can see that the red line passes through the point and that the two lines are parallel.


Submit your answer as:

ID is: 2776 Seed is: 3633

Finding the equation of a line

Determine the equation of the line which passes through the point (4;5) and is perpendicular to the line y=x31.

Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.

Answer:
equation
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. From the given equation you can see that the gradient is m=13.

For lines that are perpendicular, the gradient must be the negative reciprocal, therefore m=3.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Substitute the coordinates (4;5) and the gradient into the equation to find the value of c.

y=mx+c5=(3)(4)+c5=12+c17=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=3x+17.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (4;5) is shown with a black dot labelled with an A. You can see that the red line passes through the point and that the lines are perpendicular to each other.


Submit your answer as:

ID is: 1215 Seed is: 7474

Finding a coordinate of a point on a perpendicular line

You are given the following diagram, which shows two perpendicular lines on the Cartesian plane. Line AB has endpoints at A(2;32) and B(x;32). The other line (the dashed line) has the equation y=2x.

Calculate the missing value of x.

Answer:

x=

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The gradient of the line AB is equal to:

1gradientPL

where PL stands for Perpendicular Line.


STEP: Write down the equation of a straight line
[−1 point ⇒ 3 / 4 points left]

We need to find the equation of line AB. So we will need the equation of a straight line:

y=mx+c

STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 4 points left]

The line AB has a gradient that is equal to:

1gradientPL

where PL stands for Perpendicular Line. The gradient of the perpendicular line is 2, so:

mAB=1gradientPL=1(2)=12

Now we know that the equation of line AB is

y=x2+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known point A into the equation, to find c:

y=x2+c(32)=(2)2+cc=12

And now we have the complete equation for line AB:

y=x212

STEP: Substitute the known value for point B into the equation and solve
[−1 point ⇒ 0 / 4 points left]

We can use the equation for line AB to find the missing coordinate. To do this, substitute the known value for point B into the equation for line AB:

y=x212(32)=x212

Solving, we get:

x=4

The correct answer is x=4.


Submit your answer as:

ID is: 1215 Seed is: 7971

Finding a coordinate of a point on a perpendicular line

You are given the following diagram, which shows two perpendicular lines on the Cartesian plane. Line AB has endpoints at A(5;1) and B(x;3). The other line (the dashed line) has the equation y=2x+1.

Calculate the missing value of x.

Answer:

x=

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The gradient of the line AB is equal to:

1gradientPL

where PL stands for Perpendicular Line.


STEP: Write down the equation of a straight line
[−1 point ⇒ 3 / 4 points left]

We need to find the equation of line AB. So we will need the equation of a straight line:

y=mx+c

STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 4 points left]

The line AB has a gradient that is equal to:

1gradientPL

where PL stands for Perpendicular Line. The gradient of the perpendicular line is 2, so:

mAB=1gradientPL=1(2)=12

Now we know that the equation of line AB is

y=x2+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known point A into the equation, to find c:

y=x2+c(1)=(5)2+cc=32

And now we have the complete equation for line AB:

y=x2+32

STEP: Substitute the known value for point B into the equation and solve
[−1 point ⇒ 0 / 4 points left]

We can use the equation for line AB to find the missing coordinate. To do this, substitute the known value for point B into the equation for line AB:

y=x2+32(3)=x2+32

Solving, we get:

x=3

The correct answer is x=3.


Submit your answer as:

ID is: 1215 Seed is: 1851

Finding a coordinate of a point on a perpendicular line

You are given the following diagram, which shows two perpendicular lines on the Cartesian plane. Line AB has endpoints at A(5;4) and B(x;1). The other line (the dashed line) has the equation y=2x+52.

Calculate the missing value of x.

Answer:

x=

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The gradient of the line AB is equal to:

1gradientPL

where PL stands for Perpendicular Line.


STEP: Write down the equation of a straight line
[−1 point ⇒ 3 / 4 points left]

We need to find the equation of line AB. So we will need the equation of a straight line:

y=mx+c

STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 4 points left]

The line AB has a gradient that is equal to:

1gradientPL

where PL stands for Perpendicular Line. The gradient of the perpendicular line is 2, so:

mAB=1gradientPL=1(2)=12

Now we know that the equation of line AB is

y=x2+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known point A into the equation, to find c:

y=x2+c(4)=(5)2+cc=32

And now we have the complete equation for line AB:

y=x2+32

STEP: Substitute the known value for point B into the equation and solve
[−1 point ⇒ 0 / 4 points left]

We can use the equation for line AB to find the missing coordinate. To do this, substitute the known value for point B into the equation for line AB:

y=x2+32(1)=x2+32

Solving, we get:

x=1

The correct answer is x=1.


Submit your answer as:

ID is: 2710 Seed is: 4770

Parallel and perpendicular lines

Investigate whether these lines are parallel, perpendicular, or neither.

8x+4y3=04x+2y=3x+2
Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, we need to know the gradient of each equation. This means that we need to start by arranging the equations in standard form for linear equations: y=mx+c.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

We answer this question by comparing the gradients of the two equations, for which the equations have to be in standard form. For the first equation this is:

8x+4y3=04y=8x+314(4y)=14(8x+3)y=2x+34

The gradient of the first equation is m=2.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now we change the second equation into standard form.

4x+2y=3x+22y=x+212(2y)=12(x+2)y=x2+1

The gradient of the second equation is m=12.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

We now compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those.

  • Are the gradients equal to each other (parallel gradients)?
  • Do they have a product of 1 (perpendicular gradients)?
  • Are neither of these true?

The two gradients have a product of 1: m1×m2=(2)(12)=1.
Therefore, the lines are perpendicular. The correct response is: perpendicular.

The two equations are graphed on the Cartesian plane shown below. We can see that the two lines sit at right angles to each other: they are perpendicular.


Submit your answer as:

ID is: 2710 Seed is: 9029

Parallel and perpendicular lines

Determine whether the equations represent lines which are parallel, perpendicular, or neither.

x=2y2y=xy+10
Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, we need to know the gradient of each equation. This means that we need to start by arranging the equations in standard form for linear equations: y=mx+c.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

We answer this question by comparing the gradients of the two equations, for which the equations have to be in standard form. For the first equation this is:

x=2y22y=x212(2y)=12(x2)y=x21

The gradient of the first equation is m=12.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now we change the second equation into standard form.

y=xy+102y=x+1012(2y)=12(x+10)y=x2+5

The gradient of the second equation is m=12.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

We now compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those.

  • Are the gradients equal to each other (parallel gradients)?
  • Do they have a product of 1 (perpendicular gradients)?
  • Are neither of these true?

The equations have equal gradients, so they must be parallel lines. The correct response is: parallel.

The two equations are graphed on the Cartesian plane shown below. We can see that the two lines run parallel to each another.


Submit your answer as:

ID is: 2710 Seed is: 9619

Parallel and perpendicular lines

Are the following lines parallel, perpendicular, or neither?

3y+5=x+206y=9x+3y+2
Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, we need to know the gradient of each equation. This means that we need to start by arranging the equations in standard form for linear equations: y=mx+c.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

We answer this question by comparing the gradients of the two equations, for which the equations have to be in standard form. For the first equation this is:

3y+5=x+203y=x+1513(3y)=13(x+15)y=x3+5

The gradient of the first equation is m=13.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now we change the second equation into standard form.

6y=9x+3y+23y=9x+213(3y)=13(9x+2)y=3x+23

The gradient of the second equation is m=3.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

We now compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those.

  • Are the gradients equal to each other (parallel gradients)?
  • Do they have a product of 1 (perpendicular gradients)?
  • Are neither of these true?

For the equations in this question, the gradients are not equal, so the lines cannot be parallel. Also, the gradients are not negative reciprocals of each other, so the lines cannot be perpendicular. The correct response is: neither.

The two equations are graphed on the Cartesian plane shown below. We can see that the lines are neither parallel nor perpendicular.


Submit your answer as:

ID is: 1302 Seed is: 7167

Calculating inclination from two points

Consider the diagram below, which shows a line through the points: A(3;72) and B(3;32).

Calculate the inclination of the line, which is the angle θ. Round your answer to 2 decimal places if appropriate.

Answer: θ = °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need the inclination of the line, which is the angle θ. The inclination is related to the gradient, so start by finding the gradient of the line.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

We need to find the inclination of the line. The inclination is the angle between the line and the positive x-axis. This is shown by the angle θ in the diagram.

The inclination of a line is directly related to the line's gradient: both represent the steepness of the line. (For an explanation about the connection between the gradient and the inclination, you can read here.) So to find the inclination, we need the gradient of the line. Be aware that there is a double negative in the top of the fraction, since point A has a negative y coordinate.

m=y2y1x2x1=(32)(72)(3)(3)=32+7233=13

STEP: Calculate the inclination using the inclination formula
[−1 point ⇒ 1 / 3 points left]

Now we can calculate the inclination of the line. We do this using the following equation:

m=tanθθ=tan1(m)=tan1(13)=18.43494...°18.43°

STEP: Adjust the answer if necessary
[−1 point ⇒ 0 / 3 points left]

If the gradient is negative we must add 180° to the above value to get an obtuse angle. In this case, the gradient, and the angle we got above, is negative. (In fact, you can see in the diagram that the angle of inclination is greater than 90 degrees. For an explanation of why we must add 180° to the answer when the gradient is negative, see this information about inclination .)

θnegative18.43°+180°θ161.57°

The inclination of the line is 161.57°.


Submit your answer as:

ID is: 1302 Seed is: 3289

Calculating inclination from two points

Consider the diagram below, which shows a line through the points: A(2;4) and B(1;72).

Calculate the inclination of the line, which is the angle θ. Round your answer to 2 decimal places if appropriate.

Answer: θ = °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need the inclination of the line, which is the angle θ. The inclination is related to the gradient, so start by finding the gradient of the line.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

We need to find the inclination of the line. The inclination is the angle between the line and the positive x-axis. This is shown by the angle θ in the diagram.

The inclination of a line is directly related to the line's gradient: both represent the steepness of the line. (For an explanation about the connection between the gradient and the inclination, you can read here.) So to find the inclination, we need the gradient of the line. Be aware that there is a double negative in the bottom of the fraction, since point A has a negative x coordinate.

m=y2y1x2x1=(72)(4)(1)(2)=7241+2=52

STEP: Calculate the inclination using the inclination formula
[−1 point ⇒ 1 / 3 points left]

Now we can calculate the inclination of the line. We do this using the following equation:

m=tanθθ=tan1(m)=tan1(52)=68.19859...°68.2°

STEP: Adjust the answer if necessary
[−1 point ⇒ 0 / 3 points left]

If the gradient is negative we must add 180° to the above value to get an obtuse angle. In this case, the gradient, and the angle we got above, is negative. (In fact, you can see in the diagram that the angle of inclination is greater than 90 degrees. For an explanation of why we must add 180° to the answer when the gradient is negative, see this information about inclination .)

θnegative68.2°+180°θ111.8°

The inclination of the line is 111.8°.


Submit your answer as:

ID is: 1302 Seed is: 4081

Calculating inclination from two points

Consider the diagram below, which shows a line through the points: A(8;1) and B(6;52).

Calculate the inclination of the line, which is the angle θ. Round your answer to 2 decimal places if appropriate.

Answer: θ = °
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need the inclination of the line, which is the angle θ. The inclination is related to the gradient, so start by finding the gradient of the line.


STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

We need to find the inclination of the line. The inclination is the angle between the line and the positive x-axis. This is shown by the angle θ in the diagram.

The inclination of a line is directly related to the line's gradient: both represent the steepness of the line. (For an explanation about the connection between the gradient and the inclination, you can read here.) So to find the inclination, we need the gradient of the line. Watch out: there are double negatives in the numerator and the denominator in this gradient calculation.

m=y2y1x2x1=(52)(1)(6)(8)=52+16+8=14

STEP: Calculate the inclination using the inclination formula
[−1 point ⇒ 0 / 2 points left]

Now we can calculate the inclination of the line. We do this using the following equation:

m=tanθθ=tan1(m)=tan1(14)=14.03624...°14.04°

STEP: Adjust the answer if necessary
[−0 points ⇒ 0 / 2 points left]

If the gradient is negative we must add 180° to the above value to get an obtuse angle. In this case, the gradient, and the angle we got above, is positive.

θ14.04°

The inclination of the line is 14.04°.


Submit your answer as:

ID is: 1303 Seed is: 7152

Working with perpendicular lines

In the diagram below, the two lines are perpendicular. The equation of the dashed red line is

y=3x52

As labelled below, the solid blue line passes through the points A(3;32) and B(x;32).

Find xB, the x-coordinate of point B.

Answer: xB =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by finding the gradient of line AB. You can do this because the lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 4 / 5 points left]

We need to find the equation of the blue line, which is called line AB. That means we need the gradient and the y-intercept of line AB.

Since the two lines in the figure are perpendicular to each other, we can find the gradient of line AB from the gradient of the dashed red line. Perpendicular gradients are negative reciprocals of each other.

The gradient of the dashed red line is 3 (we can read it straight from the equation), so:

mAB=negative reciprocal of mred line=reciprocal of 3=(13)=13

STEP: Start building the equation
[−1 point ⇒ 3 / 5 points left]

Since the equation for a staight line is:

y=mx+c

The equation for line AB must be:

y=13x+c

STEP: Find the y-intercept of the equation
[−1 point ⇒ 2 / 5 points left]

Now we substitute the coordinates of point A into the equation, to find c:

y=13x+c32=13(3)+c12=c

Now we know the equation for line AB:

y=x312

STEP: Use the y-coordinate of point B to find xB
[−2 points ⇒ 0 / 5 points left]

Now we use a key concept for graphs and their equations: every point on the line agrees with the equation. If we substitute in the y-coordinate of any point on the line into the equation, we will always get the x-coordinate for that point.The equation forces this to be true! So now substitute the y-coordinate for point B into the equation for line AB and solve for xB.

yB=xB312(32)=xB312xB=6

The correct answer is xB=6.


Submit your answer as:

ID is: 1303 Seed is: 8708

Working with perpendicular lines

In the diagram below, the two lines are perpendicular. The equation of the dashed red line is

y=3x44

As labelled below, the solid blue line passes through the points A(3;5) and B(x;3).

Determine xB, the x-coordinate of point B.

Answer: xB =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by finding the gradient of line AB. You can do this because the lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 4 / 5 points left]

We need to find the equation of the blue line, which is called line AB. That means we need the gradient and the y-intercept of line AB.

Since the two lines in the figure are perpendicular to each other, we can find the gradient of line AB from the gradient of the dashed red line. Perpendicular gradients are negative reciprocals of each other.

The gradient of the dashed red line is 34 (we can read it straight from the equation), so:

mAB=negative reciprocal of mred line=reciprocal of 34=(43)=43

STEP: Start building the equation
[−1 point ⇒ 3 / 5 points left]

Since the equation for a staight line is:

y=mx+c

The equation for line AB must be:

y=43x+c

STEP: Find the y-intercept of the equation
[−1 point ⇒ 2 / 5 points left]

Now we substitute the coordinates of point A into the equation, to find c:

y=43x+c5=43(3)+c1=c

Now we know the equation for line AB:

y=4x31

STEP: Use the y-coordinate of point B to find xB
[−2 points ⇒ 0 / 5 points left]

Now we use a key concept for graphs and their equations: every point on the line agrees with the equation. If we substitute in the y-coordinate of any point on the line into the equation, we will always get the x-coordinate for that point.The equation forces this to be true! So now substitute the y-coordinate for point B into the equation for line AB and solve for xB.

yB=4xB31(3)=4xB31xB=3

The correct answer is xB=3.


Submit your answer as:

ID is: 1303 Seed is: 8412

Working with perpendicular lines

In the diagram below, the two lines are perpendicular. The equation of the dashed red line is

y=2x53

As labelled below, the solid blue line passes through the points A(1;92) and B(x;112).

Find xB, the x-coordinate of point B.

Answer: xB =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by finding the gradient of line AB. You can do this because the lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 4 / 5 points left]

We need to find the equation of the blue line, which is called line AB. That means we need the gradient and the y-intercept of line AB.

Since the two lines in the figure are perpendicular to each other, we can find the gradient of line AB from the gradient of the dashed red line. Perpendicular gradients are negative reciprocals of each other.

The gradient of the dashed red line is 25 (we can read it straight from the equation), so:

mAB=negative reciprocal of mred line=reciprocal of 25=(52)=52

STEP: Start building the equation
[−1 point ⇒ 3 / 5 points left]

Since the equation for a staight line is:

y=mx+c

The equation for line AB must be:

y=52x+c

STEP: Find the y-intercept of the equation
[−1 point ⇒ 2 / 5 points left]

Now we substitute the coordinates of point A into the equation, to find c:

y=52x+c92=52(1)+c2=c

Now we know the equation for line AB:

y=5x2+2

STEP: Use the y-coordinate of point B to find xB
[−2 points ⇒ 0 / 5 points left]

Now we use a key concept for graphs and their equations: every point on the line agrees with the equation. If we substitute in the y-coordinate of any point on the line into the equation, we will always get the x-coordinate for that point.The equation forces this to be true! So now substitute the y-coordinate for point B into the equation for line AB and solve for xB.

yB=5xB2+2(112)=5xB2+2xB=3

The correct answer is xB=3.


Submit your answer as:

ID is: 1582 Seed is: 5885

Facts about perpendicular lines

  1. From the following table, which statement or statements are true about perpendicular lines and the gradients of perpendicular lines? Identify all of the true statements.

    A m1×m2=1
    B Perpendicular gradients have a product of 1.
    C m1×m2=1
    D Perpendicular gradients are equal.
    INSTRUCTION: There may be more than one true statement above. You need to identify all true statements.
    Answer: Statement(s) is (are) true
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    There is only one true statement in the table above.
    STEP: Use facts about perpendicular lines
    [−1 point ⇒ 0 / 1 points left]

    The following statements are true for the gradients of perpendicular lines:

    • The gradients must be negative reciprocals of each other. For example, the negative reciprocal of 23 is 32.
    • The gradients must have a product of -1 : m1×m2=1. For example, 23 and 32 are perpendicular gradients because
      (23)×(32)=1
      .

    There is only one statement above which agrees with these facts.

    The correct choice is: A.


    Submit your answer as:
  2. If a certain line has a gradient of m=3, what is the gradient of a line perpendicular to that line? (Remember that division by zero is 'undefined.')

    Answer: m=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the facts from Question 1 to work out the answer!


    STEP: Find the negative reciprocal of the given gradient
    [−1 point ⇒ 0 / 1 points left]

    If the gradients are perpendicular, then the gradient we want must be the negative reciprocal of the gradient given. 'Negative' means change the sign (multiply by -1) and 'reciprocal' means turn the number upside down. The gradient given is m=3: the negative of that is 3. Then the reciprocal of that is 13.

    Note: for perpendicular gradients, m1×m2=1. The gradient given is m=3, so we want to solve (3)×m2=1. The value which will solve that equation is 13.

    Finally, we can look at a graph of lines which have the gradients above. You can see that they are perpendicular to each other! You can also see that one of the lines has a positive gradient while the other one has a negative gradient - they are negatives.

    The correct answer is 13.


    Submit your answer as:

ID is: 1582 Seed is: 9553

Facts about perpendicular lines

  1. From the following table, which statement or statements are true about perpendicular lines and the gradients of perpendicular lines? Identify all of the true statements.

    A m1×m2=1
    B Perpendicular gradients have a product of -1.
    C Perpendicular gradients have a sum of -1.
    D Perpendicular gradients are equal.
    INSTRUCTION: There may be more than one true statement above. You need to identify all true statements.
    Answer: Statement(s) is (are) true
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    There is only one true statement in the table above.
    STEP: Use facts about perpendicular lines
    [−1 point ⇒ 0 / 1 points left]

    The following statements are true for the gradients of perpendicular lines:

    • The gradients must be negative reciprocals of each other. For example, the negative reciprocal of 1 is 1.
    • The gradients must have a product of -1 : m1×m2=1. For example, 1 and 1 are perpendicular gradients because
      (1)×(1)=1
      .

    There is only one statement above which agrees with these facts.

    The correct choice is: B.


    Submit your answer as:
  2. If a certain line has a gradient of m=12, what is the gradient of a line perpendicular to that line? (Remember that division by zero is 'undefined.')

    Answer: m=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the facts from Question 1 to work out the answer!


    STEP: Find the negative reciprocal of the given gradient
    [−1 point ⇒ 0 / 1 points left]

    If the gradients are perpendicular, then the gradient we want must be the negative reciprocal of the gradient given. 'Negative' means change the sign (multiply by -1) and 'reciprocal' means turn the number upside down. The gradient given is m=12: the negative of that is 12. Then the reciprocal of that is 2.

    Note: for perpendicular gradients, m1×m2=1. The gradient given is m=12, so we want to solve (12)×m2=1. The value which will solve that equation is 2.

    Finally, we can look at a graph of lines which have the gradients above. You can see that they are perpendicular to each other! You can also see that one of the lines has a positive gradient while the other one has a negative gradient - they are negatives.

    The correct answer is 2.


    Submit your answer as:

ID is: 1582 Seed is: 717

Facts about perpendicular lines

  1. From the following table, which statement or statements are true about perpendicular lines and the gradients of perpendicular lines? Identify all of the true statements.

    A Perpendicular gradients are negative reciprocals of each other.
    B Perpendicular gradients have a product of -1.
    C Perpendicular gradients have a product of 1.
    D Perpendicular gradients are equal.
    INSTRUCTION: There may be more than one true statement above. You need to identify all true statements.
    Answer: Statement(s) is (are) true
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    There are two true statements in the table above.
    STEP: Use facts about perpendicular lines
    [−1 point ⇒ 0 / 1 points left]

    The following statements are true for the gradients of perpendicular lines:

    • The gradients must be negative reciprocals of each other. For example, the negative reciprocal of 23 is 32.
    • The gradients must have a product of -1 : m1×m2=1. For example, 23 and 32 are perpendicular gradients because
      (23)×(32)=1
      .

    There are two statements in the table above which agree with these facts.

    The correct choice is: A and B.


    Submit your answer as:
  2. If a certain line has a gradient of m=13, what is the gradient of a line perpendicular to that line? (Remember that division by zero is 'undefined.')

    Answer: m=
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the facts from Question 1 to work out the answer!


    STEP: Find the negative reciprocal of the given gradient
    [−1 point ⇒ 0 / 1 points left]

    If the gradients are perpendicular, then the gradient we want must be the negative reciprocal of the gradient given. 'Negative' means change the sign (multiply by -1) and 'reciprocal' means turn the number upside down. The gradient given is m=13: the negative of that is 13. Then the reciprocal of that is 3.

    Note: for perpendicular gradients, m1×m2=1. The gradient given is m=13, so we want to solve (13)×m2=1. The value which will solve that equation is 3.

    Finally, we can look at a graph of lines which have the gradients above. You can see that they are perpendicular to each other! You can also see that one of the lines has a positive gradient while the other one has a negative gradient - they are negatives.

    The correct answer is 3.


    Submit your answer as:

ID is: 1298 Seed is: 4827

Finding the equation of a perpendicular line

The figure below shows two perpendicular lines. The equation of the dashed red line is:

y=2x+32

The solid blue line passes through points A and B. The coordinates of point A are (7;3) and the coordinates of B are unknown.

Calculate the equation of line AB.

INSTRUCTION: Type only the right side of the equation into the answer box (following the "y = ...").
Answer: Equation of line AB: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the gradient of line AB. You can do this using the fact that the two lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

The equation of a straight line requires the gradient and the y-intercept for the line. We can find the gradient in this case because line AB is perpendicular to the dashed red line: if we know a gradient, we can always find the gradient which is perpendicular to it.

The equation of the dashed red line is

y=2x+32

which means that the gradient of the dashed line is 2.

Perpendicular lines have gradients which are negative reciprocals of each other. So the gradient of line AB is the negative reciprocal of 2.

mAB=negative reciprocal of mdashed line=reciprocal of 2=(12)=12

This means that the equation of line AB is:

y=x2+c

STEP: Find the value of c
[−1 point ⇒ 1 / 3 points left]

Now we can find the value of c for the line (which is the y-intercept). We do this by substituting any point on the line into the equation from above. We only know the coordinates of point A, which are (7;3). Substitute these coordinates into the equation and solve for c.

y=x2+c(3)=(7)2+ccoordinates of point ASubstitute in the12=c

STEP: Write the equation
[−1 point ⇒ 0 / 3 points left]

We now have both numbers we need for the equation of line AB: we just need to put them together.

The correct answer is:

y=x212

Submit your answer as:

ID is: 1298 Seed is: 6893

Finding the equation of a perpendicular line

The figure below shows two perpendicular lines. The equation of the dashed red line is:

y=x2+32

The solid blue line passes through points A and B. The coordinates of point A are (3;92) and the coordinates of B are unknown.

Calculate the equation of line AB.

INSTRUCTION: Type only the right side of the equation into the answer box (following the "y = ...").
Answer: Equation of line AB: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the gradient of line AB. You can do this using the fact that the two lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

The equation of a straight line requires the gradient and the y-intercept for the line. We can find the gradient in this case because line AB is perpendicular to the dashed red line: if we know a gradient, we can always find the gradient which is perpendicular to it.

The equation of the dashed red line is

y=x2+32

which means that the gradient of the dashed line is 12.

Perpendicular lines have gradients which are negative reciprocals of each other. So the gradient of line AB is the negative reciprocal of 12.

mAB=negative reciprocal of mdashed line=reciprocal of 12=(2)=2

This means that the equation of line AB is:

y=2x+c

STEP: Find the value of c
[−1 point ⇒ 1 / 3 points left]

Now we can find the value of c for the line (which is the y-intercept). We do this by substituting any point on the line into the equation from above. We only know the coordinates of point A, which are (3;92). Substitute these coordinates into the equation and solve for c.

y=2x+c(92)=2(3)+ccoordinates of point ASubstitute in the32=c

STEP: Write the equation
[−1 point ⇒ 0 / 3 points left]

We now have both numbers we need for the equation of line AB: we just need to put them together.

The correct answer is:

y=2x32

Submit your answer as:

ID is: 1298 Seed is: 4160

Finding the equation of a perpendicular line

The figure below shows two perpendicular lines. The equation of the dashed red line is:

y=3x1

The solid blue line passes through points A and B. The coordinates of point A are (4;53) and the coordinates of B are unknown.

Determine the equation of line AB.

INSTRUCTION: Type only the right side of the equation into the answer box (following the "y = ...").
Answer: Equation of line AB: y =
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the gradient of line AB. You can do this using the fact that the two lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

The equation of a straight line requires the gradient and the y-intercept for the line. We can find the gradient in this case because line AB is perpendicular to the dashed red line: if we know a gradient, we can always find the gradient which is perpendicular to it.

The equation of the dashed red line is

y=3x1

which means that the gradient of the dashed line is 3.

Perpendicular lines have gradients which are negative reciprocals of each other. So the gradient of line AB is the negative reciprocal of 3.

mAB=negative reciprocal of mdashed line=reciprocal of 3=(13)=13

This means that the equation of line AB is:

y=x3+c

STEP: Find the value of c
[−1 point ⇒ 1 / 3 points left]

Now we can find the value of c for the line (which is the y-intercept). We do this by substituting any point on the line into the equation from above. We only know the coordinates of point A, which are (4;53). Substitute these coordinates into the equation and solve for c.

y=x3+c(53)=(4)3+ccoordinates of point ASubstitute in the3=c

STEP: Write the equation
[−1 point ⇒ 0 / 3 points left]

We now have both numbers we need for the equation of line AB: we just need to put them together.

The correct answer is:

y=x33

Submit your answer as:

ID is: 2693 Seed is: 5200

Finding a point using perpendicular lines

The following diagram is given:

Line AB is perpendicular to the line y=2x0,25.

Find the y-coordinate of point B, rounded to 2 decimal places.

Answer: B (2;)
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Recall the standard form of a straight line equation:

y=mx+c

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

For perpendicular lines, we know that m1×m2=1:

mAB×2=1mAB=12=0.5y=0.5x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

We substitute the x and y coordinates of point A into the equation to find c:

y=0.5x+c1=(0.5)(1)+cc=0.5

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Finally, we substitute the x-coordinate for point B into the equation and solve for y:

y=0.5x0.5y=(0.5)(2)0.5y=0.5

Submit your answer as:

ID is: 2693 Seed is: 2487

Finding a point using perpendicular lines

The following diagram is given:

Line AB is perpendicular to the line y=x.

Find the y-coordinate of point B, rounded to 2 decimal places.

Answer: B (4;)
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Recall the standard form of a straight line equation:

y=mx+c

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

For perpendicular lines, we know that m1×m2=1:

mAB×1=1mAB=11=1y=1x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

We substitute the x and y coordinates of point A into the equation to find c:

y=1x+c3.5=(1)(3)+cc=0.5

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Finally, we substitute the x-coordinate for point B into the equation and solve for y:

y=1x0.5y=(1)(4)0.5y=3.5

Submit your answer as:

ID is: 2693 Seed is: 6460

Finding a point using perpendicular lines

The following diagram is given:

Line AB is perpendicular to the line y=2x1,75.

Calculate the x-coordinate of point B, rounded to 2 decimal places.

Answer: B (;4)
numeric
2 attempts remaining
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Recall the standard form of a straight line equation:

y=mx+c

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

For perpendicular lines, we know that m1×m2=1:

mAB×2=1mAB=12=0.5y=0.5x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

We substitute the x and y coordinates of point A into the equation to find c:

y=0.5x+c0.5=(0.5)(5)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Finally, we substitute the y-coordinate for point B into the equation and solve for x:

y=0.5x24=(0.5)(x)2x=4

Submit your answer as:

ID is: 1301 Seed is: 9383

Finding the equation of a line from the inclination

The diagram below shows a line on the Cartesian plane. The line passes through the point A(4,5). The line has an angle of inclination θ=63.435°.

Determine the equation of the line.

INSTRUCTIONS:
  • Give only the right side of the equation, following y=....
  • Round the values in the equation to 2 decimal places if necessary.
Answer: Complete the equation for the line: y = .
one-of
type(expression)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the inclination to find the gradient of the line. Then use that gradient to start working out the equation.


STEP: Use the inclination to find the gradient of the line
[−1 point ⇒ 3 / 4 points left]

This question is about finding the equation of a straight line. We know one point on the line and the line's inclination. The equation of the line will have the form y=mx+c. So we need to find the gradient, m, and the y-intercept, c.

The inclination of a line tells us about the steepness of the line so it is similar to the gradient. In fact, inclination and gradient are directly related to each other. We can use the following equation to calculate the gradient from the inclination.

m=tanθ=tan(63.435°)=2.00000...2

The gradient is approximately 2 (remember that the instructions state that we should round the values to 2 decimal places). Note that the gradient is positive, which agrees with the graph.


STEP: Put the gradient into the equation
[−1 point ⇒ 2 / 4 points left]

Now we recall the general form of the equation of a straight line. Since we know the gradient m of the line, we can substitute that into the equation of the line:

y=(2)x+c

STEP: Use the point to find the y-intercept
[−1 point ⇒ 1 / 4 points left]

Now we can calculate the value of the y-intercept of the line (the value of c). We do this by substituting the coordinates of any point on the line into the general form for the line from above. Since we know the coordinates of point A, we will use those coordinates, which are (4,5).

y=2x+c(5)=2(4)+c3=c

Remember that this number should be the y-intercept of the graph. As with the gradient, we can compare this answer to the graph to check if they agree: does the graph up above have a y-intercept close to 3? Yes!


STEP: Write the complete equation
[−1 point ⇒ 0 / 4 points left]

We now have the gradient and the y-intercept of the line. The equation of the line is:

y=2x3

Submit your answer as:

ID is: 1301 Seed is: 4413

Finding the equation of a line from the inclination

The diagram below shows a line on the Cartesian plane. The line passes through the point A(3,114). The line has an angle of inclination θ=128.66°.

Determine the equation of the line.

INSTRUCTIONS:
  • Give only the right side of the equation, following y=....
  • Round the values in the equation to 2 decimal places if necessary.
Answer: Complete the equation for the line: y = .
one-of
type(expression)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the inclination to find the gradient of the line. Then use that gradient to start working out the equation.


STEP: Use the inclination to find the gradient of the line
[−1 point ⇒ 3 / 4 points left]

This question is about finding the equation of a straight line. We know one point on the line and the line's inclination. The equation of the line will have the form y=mx+c. So we need to find the gradient, m, and the y-intercept, c.

The inclination of a line tells us about the steepness of the line so it is similar to the gradient. In fact, inclination and gradient are directly related to each other. We can use the following equation to calculate the gradient from the inclination.

m=tanθ=tan(128.66°)=1.24999...1.25

The gradient is approximately 1.25 (remember that the instructions state that we should round the values to 2 decimal places). Note that the gradient is negative, which agrees with the slope of the graph.


STEP: Put the gradient into the equation
[−1 point ⇒ 2 / 4 points left]

Now we recall the general form of the equation of a straight line. Since we know the gradient m of the line, we can substitute that into the equation of the line:

y=(1.25)x+c

STEP: Use the point to find the y-intercept
[−1 point ⇒ 1 / 4 points left]

Now we can calculate the value of the y-intercept of the line (the value of c). We do this by substituting the coordinates of any point on the line into the general form for the line from above. Since we know the coordinates of point A, we will use those coordinates, which are (3,114).

y=1.25x+c(114)=1.25(3)+c1=c

Remember that this number should be the y-intercept of the graph. As with the gradient, we can compare this answer to the graph to check if they agree: does the graph up above have a y-intercept close to 1? Yes!


STEP: Write the complete equation
[−1 point ⇒ 0 / 4 points left]

We now have the gradient and the y-intercept of the line. The equation of the line is:

y=1.25x+1

Submit your answer as:

ID is: 1301 Seed is: 8666

Finding the equation of a line from the inclination

The diagram below shows a line on the Cartesian plane. The line passes through the point A(2,92). The line has an angle of inclination θ=71.565°.

Find the equation of the line.

INSTRUCTIONS:
  • Give only the right side of the equation, following y=....
  • Round the values in the equation to 2 decimal places if necessary.
Answer: Complete the equation for the line: y = .
one-of
type(expression)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the inclination to find the gradient of the line. Then use that gradient to start working out the equation.


STEP: Use the inclination to find the gradient of the line
[−1 point ⇒ 3 / 4 points left]

This question is about finding the equation of a straight line. We know one point on the line and the line's inclination. The equation of the line will have the form y=mx+c. So we need to find the gradient, m, and the y-intercept, c.

The inclination of a line tells us about the steepness of the line so it is similar to the gradient. In fact, inclination and gradient are directly related to each other. We can use the following equation to calculate the gradient from the inclination.

m=tanθ=tan(71.565°)=2.99999...3

The gradient is approximately 3 (remember that the instructions state that we should round the values to 2 decimal places). Note that the gradient is positive, which agrees with the graph.


STEP: Put the gradient into the equation
[−1 point ⇒ 2 / 4 points left]

Now we recall the general form of the equation of a straight line. Since we know the gradient m of the line, we can substitute that into the equation of the line:

y=(3)x+c

STEP: Use the point to find the y-intercept
[−1 point ⇒ 1 / 4 points left]

Now we can calculate the value of the y-intercept of the line (the value of c). We do this by substituting the coordinates of any point on the line into the general form for the line from above. Since we know the coordinates of point A, we will use those coordinates, which are (2,92).

y=3x+c(92)=3(2)+c1.5=c

Remember that this number should be the y-intercept of the graph. As with the gradient, we can compare this answer to the graph to check if they agree: does the graph up above have a y-intercept close to 1.5? Yes!


STEP: Write the complete equation
[−1 point ⇒ 0 / 4 points left]

We now have the gradient and the y-intercept of the line. The equation of the line is:

y=3x1.5

Submit your answer as:

6. Practical application


ID is: 4403 Seed is: 4042

Equation of straight line & point of intersection

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight line PQ is drawn with P(0,83) and Q(4,0) as its y- and x- intercepts respectively. The equation of PQ is 23x+y83=0. The straight line SR is drawn with S on PQ and SRQP. SR cuts the x-axis at R(52,0).

  1. The equation of Line SR can be written in the form y=mx+c. Determine the values of m and c.

    INSTRUCTION: Where necessary, give your answer as a fraction.
    Answer:
    • m=
    • c=
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    QPSR therefore mQP×mSR=1.


    STEP: Determine the gradients of QP and SR
    [−2 points ⇒ 1 / 3 points left]
    TIP: Perpendicular lines have negative inverse gradients.
    mQP=yQyPxQxP=(0)(83)(4)(0)=23

    QPSR therefore mQP×mSR=1 i.e. 23×mSR=1

    mSR=32
    NOTE: You could have calculated mQP by writing the equation for PQ in standard form (since the m-value in y=mx+c is the gradient).

    STEP: R is the x-intercept of SR
    [−1 point ⇒ 0 / 3 points left]

    The equation of SR is given by y=32x+c.

    The x-intercept is R(52,0). Substitute this into the equation of SR:

    y=32x+c(0)=32(52)+cc=154

    The equation of SR is y=32x+154.

    Therefore m=32 and c=154.


    Submit your answer as: and
  2. Determine the coordinates of S, the point where PQ and SR meet.

    INSTRUCTION:

    Type brackets around the x- and y-values and type a semicolon in between like this: (x; y)

    Answer:

    Coordinates of S:

    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    S is the point where PQ and SR intersect, hence the equation of PQ is equal to the equation of SR at this point.


    STEP: Solve simultaneous equations
    [−4 points ⇒ 0 / 4 points left]
    TIP: At the point where two lines intersect the equation of the one line is equal to the equation of the other line.

    The equation of PQ was given as 23x+y83=0.

    We know that the equation of SR is: y=32x+154.

    S is the point where PQ and SR intersect, hence the equation of PQ is equal to the equation of SR at this point. Since the equations are not both in standard form, we will use the substitution method.

    23x+(32x+154)=83136x=1312x=12

    Now, to find the y-value when x=12 substitute into the equation of SR.

    y=32(12)+154y=3

    So, S is the point where x=12 and y=3.

    The coordinate S is (12,3).


    Submit your answer as:

ID is: 4403 Seed is: 5384

Equation of straight line & point of intersection

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight line KL is drawn with K(0,152) and L(5,0) as its y- and x- intercepts respectively. The equation of KL is 32x+y152=0. The straight line NM is drawn with N on KL and NMLK. NM cuts the x-axis at M(32,0).

  1. The equation of Line NM can be written in the form y=mx+c. Determine the values of m and c.

    INSTRUCTION: Where necessary, give your answer as a fraction.
    Answer:
    • m=
    • c=
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    LKNM therefore mLK×mNM=1.


    STEP: Determine the gradients of LK and NM
    [−2 points ⇒ 1 / 3 points left]
    TIP: Perpendicular lines have negative inverse gradients.
    mLK=yLyKxLxK=(0)(152)(5)(0)=32

    LKNM therefore mLK×mNM=1 i.e. 32×mNM=1

    mNM=23
    NOTE: You could have calculated mLK by writing the equation for KL in standard form (since the m-value in y=mx+c is the gradient).

    STEP: M is the x-intercept of NM
    [−1 point ⇒ 0 / 3 points left]

    The equation of NM is given by y=23x+c.

    The x-intercept is M(32,0). Substitute this into the equation of NM:

    y=23x+c(0)=23(32)+cc=1

    The equation of NM is y=23x+1.

    Therefore m=23 and c=1.


    Submit your answer as: and
  2. Determine the coordinates of N, the point where KL and NM meet.

    INSTRUCTION:

    Type brackets around the x- and y-values and type a semicolon in between like this: (x; y)

    Answer:

    Coordinates of N:

    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    N is the point where KL and NM intersect, hence the equation of KL is equal to the equation of NM at this point.


    STEP: Solve simultaneous equations
    [−4 points ⇒ 0 / 4 points left]
    TIP: At the point where two lines intersect the equation of the one line is equal to the equation of the other line.

    The equation of KL was given as 32x+y152=0.

    We know that the equation of NM is: y=23x+1.

    N is the point where KL and NM intersect, hence the equation of KL is equal to the equation of NM at this point. Since the equations are not both in standard form, we will use the substitution method.

    32x+(23x+1)=152136x=132x=3

    Now, to find the y-value when x=3 substitute into the equation of NM.

    y=23(3)+1y=3

    So, N is the point where x=3 and y=3.

    The coordinate N is (3,3).


    Submit your answer as:

ID is: 4403 Seed is: 2979

Equation of straight line & point of intersection

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight line KL is drawn with K(6,0) and L(0,4) as its x- and y- intercepts respectively. The equation of KL is x32y+6=0. The straight line NM is drawn with N on KL and NMLK. NM cuts the y-axis at M(0,52).

  1. The equation of Line NM can be written in the form y=mx+c. Determine the values of m and c.

    INSTRUCTION: Where necessary, give your answer as a fraction.
    Answer:
    • m=
    • c=
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    LKNM therefore mLK×mNM=1.


    STEP: Determine the gradients of LK and NM
    [−2 points ⇒ 1 / 3 points left]
    TIP: Perpendicular lines have negative inverse gradients.
    mLK=yLyKxLxK=(4)(0)(0)(6)=23

    LKNM therefore mLK×mNM=1 i.e. 23×mNM=1

    mNM=32
    NOTE: You could have calculated mLK by writing the equation for KL in standard form (since the m-value in y=mx+c is the gradient).

    STEP: M is the y-intercept of NM
    [−1 point ⇒ 0 / 3 points left]
    The constant, c, in the standard form equation of a straight line y=mx+c is equal to the y-intercept of the line.

    The y-intercept is M(0,52)

    c=52

    The equation of NM is y=32x52.

    Therefore m=32 and c=52.


    Submit your answer as: and
  2. Determine the coordinates of N, the point where KL and NM meet.

    INSTRUCTION:

    Type brackets around the x- and y-values and type a semicolon in between like this: (x; y)

    Answer:

    Coordinates of N:

    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    N is the point where KL and NM intersect, hence the equation of KL is equal to the equation of NM at this point.


    STEP: Solve simultaneous equations
    [−4 points ⇒ 0 / 4 points left]
    TIP: At the point where two lines intersect the equation of the one line is equal to the equation of the other line.

    The equation of KL was given as x32y+6=0.

    We know that the equation of NM is: y=32x52.

    N is the point where KL and NM intersect, hence the equation of KL is equal to the equation of NM at this point. Since the equations are not both in standard form, we will use the substitution method.

    x32(32x52)=6x+94x+154=6134x=394x=3

    Now, to find the y-value when x=3 substitute into the equation of NM.

    y=32(3)52y=2

    So, N is the point where x=3 and y=2.

    The coordinate N is (3,2).


    Submit your answer as:

ID is: 4404 Seed is: 3884

Straight line application

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight line DE is drawn with D and E as its y- and x- intercepts respectively. The equation of DE is bx+y5b=0 where b0. It is also given that 2OD=3OE. The straight line GF is drawn with G on DE and GFED. GF cuts the x-axis at F(32,0).

  1. Calculate the coordinates of E.

    INSTRUCTION:
    • Type brackets around the x- and y-values and type a semicolon in between like this: (x; y)
    • Type the x- and y-values as simplified fractions, if necessary.
    Answer: Coordinates of E:
    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    E is the x-intercept of DE. Substitute y=0 into the equation of DE.


    STEP: Substitute y=0 into the equation of DE
    [−2 points ⇒ 0 / 2 points left]

    We are given the equation of DE: bx+y5b=0. E is the x-intercept of DE, so at this point y=0.

    bx+y5b=0bx+(0)5b=0b(x5)=0andb0x=5

    Therefore the coordinates of E are: E(5,0).

    NOTE: You could instead find the coordinates of E by choosing to write the equation of DE in standard form (since the c-value in y=mx+c is the y-intercept).

    Submit your answer as:
  2. Calculate the value of b.

    INSTRUCTION: Where necessary, give your answer as a simplified fraction.
    Answer: b=
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The x-value of E gives the distance OE, and 2OD=3OE.


    STEP: Determine the coordinates of D
    [−1 point ⇒ 1 / 2 points left]

    The x-value of E(5,0) gives the distance OE , so OE is 5 units.

    We are given 2OD=3OE, so OD=5 units×3÷2 which is 152 units.

    The y-value of D is the distance 152.

    Therefore the point D is: (0,152).


    STEP: Use the fact that D is a point on the line DE
    [−1 point ⇒ 0 / 2 points left]

    Substitute D(0,152) into the equation of DE.

    bx+y5b=0b(0)+(152)5b=032=b
    NOTE: You could instead work out the gradient of ED and use this to determine the value of b.

    Submit your answer as:

ID is: 4404 Seed is: 5155

Straight line application

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight line SP is drawn with S and P as its y- and x- intercepts respectively. The equation of SP is nx+yn=0 where n0. It is also given that 3OS=OP. The straight line RT is drawn with R on SP and RTPS. RT cuts the x-axis at T(173,0).

  1. Calculate the coordinates of P.

    INSTRUCTION:
    • Type brackets around the x- and y-values and type a semicolon in between like this: (x; y)
    • Type the x- and y-values as simplified fractions, if necessary.
    Answer: Coordinates of P:
    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    P is the x-intercept of SP. Substitute y=0 into the equation of SP.


    STEP: Substitute y=0 into the equation of SP
    [−2 points ⇒ 0 / 2 points left]

    We are given the equation of SP: nx+yn=0. P is the x-intercept of SP, so at this point y=0.

    nx+yn=0nx+(0)n=0n(x1)=0andn0x=1

    Therefore the coordinates of P are: P(1,0).

    NOTE: You could instead find the coordinates of P by choosing to write the equation of SP in standard form (since the c-value in y=mx+c is the y-intercept).

    Submit your answer as:
  2. Calculate the value of n.

    INSTRUCTION: Where necessary, give your answer as a simplified fraction.
    Answer: n=
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The x-value of P gives the distance OP, and 3OS=OP.


    STEP: Determine the coordinates of S
    [−1 point ⇒ 1 / 2 points left]

    The x-value of P(1,0) gives the distance OP , so OP is 1 unit.

    We are given 3OS=OP, so OS=1 unit×1÷3 which is 13 units.

    The y-value of S is the distance 13.

    Therefore the point S is: (0,13).


    STEP: Use the fact that S is a point on the line SP
    [−1 point ⇒ 0 / 2 points left]

    Substitute S(0,13) into the equation of SP.

    nx+yn=0n(0)+(13)n=013=n
    NOTE: You could instead work out the gradient of PS and use this to determine the value of n.

    Submit your answer as:

ID is: 4404 Seed is: 9162

Straight line application

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight line PQ is drawn with P and Q as its x- and y- intercepts respectively. The equation of PQ is xpy+5p=0 where p0. It is also given that 2OP=3OQ. The straight line SR is drawn with S on PQ and SRQP. SR cuts the y-axis at R(0,32).

  1. Calculate the coordinates of Q.

    INSTRUCTION:
    • Type brackets around the x- and y-values and type a semicolon in between like this: (x; y)
    • Type the x- and y-values as simplified fractions, if necessary.
    Answer: Coordinates of Q:
    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Q is the y-intercept of PQ. Substitute x=0 into the equation of PQ.


    STEP: Substitute x=0 into the equation of PQ
    [−2 points ⇒ 0 / 2 points left]

    We are given the equation of PQ: xpy+5p=0. Q is the y-intercept of PQ, so at this point x=0.

    xpy+5p=0(0)py+5p=0p(y+5)=0andp0y=5

    Therefore the coordinates of Q are: Q(0,5).

    NOTE: You could instead find the coordinates of Q by choosing to write the equation of PQ in standard form (since the c-value in y=mx+c is the y-intercept).

    Submit your answer as:
  2. Calculate the value of p.

    INSTRUCTION: Where necessary, give your answer as a simplified fraction.
    Answer: p=
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The y-value of Q gives the distance OQ, and 2OP=3OQ.


    STEP: Determine the coordinates of P
    [−1 point ⇒ 1 / 2 points left]

    The y-value of Q(0,5) gives the distance OQ , so OQ is 5 units.

    We are given 2OP=3OQ, so OP=5 units×3÷2 which is 152 units.

    But we can see from the diagram that the x-value of P must be negative. So the x-value of P is 152.

    Therefore the point P is: (152,0).


    STEP: Use the fact that P is a point on the line PQ
    [−1 point ⇒ 0 / 2 points left]

    Substitute P(152,0) into the equation of PQ.

    xpy+5p=0(152)p(0)+5p=032=p
    NOTE: You could instead work out the gradient of QP and use this to determine the value of p.

    Submit your answer as:

ID is: 4391 Seed is: 1383

Analytical geometry: mixed applications

Adapted from DBE Nov 2015 Grade 11, P2, Q3
Maths formulas

In the diagram, P(5,3), R(16,15) and S(17,2) are the vertices of ΔPRS. RP intersects the y-axis at U and RP produced meets the x-axis at V. The straight line ST is drawn parallel to RP. RUO^=α.

  1. Calculate the gradient of the line PR.

    INSTRUCTION: Give your answer as a fraction.
    Answer: Gradient of PR= .
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Gradient formula: m=y2y1x2x1.


    STEP: Use the gradient formula
    [−2 points ⇒ 0 / 2 points left]

    m=y2y1x2x1.

    mPR=15(3)165=47

    Submit your answer as:
  2. The equation of Line ST can be written in the form y=mx+c. Determine the values of m and c.

    Here is the diagram again:

    Answer:
    • m=
    • c=
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Parallel lines have the same gradient.


    STEP: Use the parallel lines to determine the gradient of ST
    [−1 point ⇒ 2 / 3 points left]
    Parallel lines have the same gradient.

    In Question 1 we found that mPR=47. Since STPR:

    mST=mPRmST=47

    So y=47x+c.


    STEP: Substitute in a point on ST and solve for c
    [−2 points ⇒ 0 / 3 points left]

    Point S(17,2) lies on ST, so we can substitute in its coordinates:

    2=47(17)+cc=827y=47x+827

    Submit your answer as: and
  3. Calculate the size of α.

    INSTRUCTION: Round your answer to two decimal places.

    Here is the diagram again:

    Answer: α= °
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Determine the relationship between α and an angle of inclination that you can calculate.


    STEP: Determine the relationship between α and the angle of inclination
    [−1 point ⇒ 2 / 3 points left]
    The angle of inclination of a line is the angle between the x-axis and the line (measured in an anti-clockwise direction from the positive x-axis).

    The acute angle θ is the angle of inclination of VR.

    Once we know the angle of inclination of VR we will be able to determine α using Triangle ΔUVO.


    STEP: Determine the angle of inclination of VR
    [−1 point ⇒ 1 / 3 points left]
    The angle of inclination is linked to gradient by the formula m=tanθ.

    From Question 1, we know that mVR=mPR=47.

    mVR=tanθ47=tanθθ=29.74488...29.74°

    STEP: Determine α
    [−1 point ⇒ 0 / 3 points left]
    UVO^=29.74°(vert opp s equal)α=90°+29.74°(ext  of Δ)=119.74°
    TIP: There are other ways that you can calculate α. As long as you use geometry theorems correctly, you should get the same answer.

    Submit your answer as:

ID is: 4391 Seed is: 7494

Analytical geometry: mixed applications

Adapted from DBE Nov 2015 Grade 11, P2, Q3
Maths formulas

In the diagram, P(5,5), R(5,11) and S(9,5) are the vertices of ΔPRS. RP intersects the y-axis at U and RP produced meets the x-axis at V. The straight line ST is drawn parallel to RP. RUO^=α.

  1. Calculate the gradient of the line PR.

    INSTRUCTION: Give your answer as a fraction.
    Answer: Gradient of PR= .
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Gradient formula: m=y2y1x2x1.


    STEP: Use the gradient formula
    [−2 points ⇒ 0 / 2 points left]

    m=y2y1x2x1.

    mPR=1155(5)=35

    Submit your answer as:
  2. The equation of Line ST can be written in the form y=mx+c. Determine the values of m and c.

    Here is the diagram again:

    Answer:
    • m=
    • c=
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Parallel lines have the same gradient.


    STEP: Use the parallel lines to determine the gradient of ST
    [−1 point ⇒ 2 / 3 points left]
    Parallel lines have the same gradient.

    In Question 1 we found that mPR=35. Since STPR:

    mST=mPRmST=35

    So y=35x+c.


    STEP: Substitute in a point on ST and solve for c
    [−2 points ⇒ 0 / 3 points left]

    Point S(9,5) lies on ST, so we can substitute in its coordinates:

    5=35(9)+cc=525y=35x525

    Submit your answer as: and
  3. Calculate the size of α.

    INSTRUCTION: Round your answer to two decimal places.

    Here is the diagram again:

    Answer: α= °
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Determine the relationship between α and an angle of inclination that you can calculate.


    STEP: Determine the relationship between α and the angle of inclination
    [−1 point ⇒ 2 / 3 points left]
    The angle of inclination of a line is the angle between the x-axis and the line (measured in an anti-clockwise direction from the positive x-axis).

    The acute angle θ is the angle of inclination of VR.

    Once we know the angle of inclination of VR we will be able to determine α using Triangle ΔUVO.


    STEP: Determine the angle of inclination of VR
    [−1 point ⇒ 1 / 3 points left]
    The angle of inclination is linked to gradient by the formula m=tanθ.

    From Question 1, we know that mVR=mPR=35.

    mVR=tanθ35=tanθθ=30.96375...30.96°

    STEP: Determine α
    [−1 point ⇒ 0 / 3 points left]
    α=90°+30.96°(ext  of Δ)=120.96°
    TIP: There are other ways that you can calculate α. As long as you use geometry theorems correctly, you should get the same answer.

    Submit your answer as:

ID is: 4391 Seed is: 3967

Analytical geometry: mixed applications

Adapted from DBE Nov 2015 Grade 11, P2, Q3
Maths formulas

In the diagram, P(5,3), R(13,15) and S(15,2) are the vertices of ΔPRS. RP intersects the y-axis at U and RP produced meets the x-axis at V. The straight line ST is drawn parallel to RP. RUO^=β.

  1. Calculate the gradient of the line PR.

    INSTRUCTION: Give your answer as a fraction.
    Answer: Gradient of PR= .
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Gradient formula: m=y2y1x2x1.


    STEP: Use the gradient formula
    [−2 points ⇒ 0 / 2 points left]

    m=y2y1x2x1.

    mPR=15313(5)=23

    Submit your answer as:
  2. The equation of Line ST can be written in the form y=mx+c. Determine the values of m and c.

    Here is the diagram again:

    Answer:
    • m=
    • c=
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Parallel lines have the same gradient.


    STEP: Use the parallel lines to determine the gradient of ST
    [−1 point ⇒ 2 / 3 points left]
    Parallel lines have the same gradient.

    In Question 1 we found that mPR=23. Since STPR:

    mST=mPRmST=23

    So y=23x+c.


    STEP: Substitute in a point on ST and solve for c
    [−2 points ⇒ 0 / 3 points left]

    Point S(15,2) lies on ST, so we can substitute in its coordinates:

    2=23(15)+cc=12y=23x12

    Submit your answer as: and
  3. Calculate the size of β.

    INSTRUCTION: Round your answer to two decimal places.

    Here is the diagram again:

    Answer: β= °
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Determine the relationship between β and an angle of inclination that you can calculate.


    STEP: Determine the relationship between β and the angle of inclination
    [−1 point ⇒ 2 / 3 points left]
    The angle of inclination of a line is the angle between the x-axis and the line (measured in an anti-clockwise direction from the positive x-axis).

    The acute angle θ is the angle of inclination of VR.

    Once we know the angle of inclination of VR we will be able to determine β using Triangle ΔUVO.


    STEP: Determine the angle of inclination of VR
    [−1 point ⇒ 1 / 3 points left]
    The angle of inclination is linked to gradient by the formula m=tanθ.

    From Question 1, we know that mVR=mPR=23.

    mVR=tanθ23=tanθθ=33.69006...33.69°

    STEP: Determine β
    [−1 point ⇒ 0 / 3 points left]
    β=90°+33.69°(ext  of Δ)=123.69°
    TIP: There are other ways that you can calculate β. As long as you use geometry theorems correctly, you should get the same answer.

    Submit your answer as:

ID is: 4402 Seed is: 9874

Analytical geometry: area of triangle and circle geometry application

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight lines EG and GF are drawn with E(4,0), F(52,0) and G(12,3) such that EGGF.

  1. Calculate the area of ΔEGF.

    INSTRUCTION: Give your answer as a simplified fraction if necessary.
    Answer:

    Area of ΔEGF= units2

    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Consider EF as the base of the triangle. What is the perpendicular height?


    STEP: Use EF and the y-value of G to calculate the area of ΔEGF
    [−3 points ⇒ 0 / 3 points left]
    The area of a triangle is equal to half of its base multiplied by the perpendicular height of the triangle: 12×b×h.

    The question becomes a lot easier if we think about EF as the base of the triangle.

    With EF as the base, the perpendicular height is the vertical line from the x-axis to the point G. In other words, we can get the height from the y-value of G, which is 3 units.

    • EF is a distance of 4(52)=132 units.
    • height is a distance of 3 units.
    Area ΔEGF=12×(132)×(3)=394 units2
    TIP: If we consider EF as the base of the triangle, EG is not the perpendicular height! The perpendicular height must be the line, perpendicular to the base EF, joining the base to the opposite vertex G.

    Submit your answer as:
  2. Calculate, giving reasons, the radius of a circle passing through the points E, F and G.

    INSTRUCTION: Give your answer as a simplified fraction or as a decimal rounded to two places if necessary.
    Answer:

    The diameter of the circle passing through E, G and F is because .

    radius = units.

    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What do you know about a chord that subtends 90° at the circumference of a circle?


    STEP: The radius is always half the diameter
    [−2 points ⇒ 0 / 2 points left]
    A chord that subtends an angle of 90° at the circumference of a circle is a diameter of the circle: chord subtends 90°.

    EF subtends 90° at the circumference of the circle passing through E, F and G.

    EF is the diameter (chord subtends 90°)

    The radius of a circle is always half the diameter.

    r=12×EF=12×132=134 units

    Submit your answer as: andand

ID is: 4402 Seed is: 5183

Analytical geometry: area of triangle and circle geometry application

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight lines PR and RT are drawn with P(52,0), T(154,0) and R(32,3) such that PRRT.

  1. Calculate the area of ΔPRT.

    INSTRUCTION: Give your answer as a simplified fraction if necessary.
    Answer:

    Area of ΔPRT= units2

    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Consider PT as the base of the triangle. What is the perpendicular height?


    STEP: Use PT and the y-value of R to calculate the area of ΔPRT
    [−3 points ⇒ 0 / 3 points left]
    The area of a triangle is equal to half of its base multiplied by the perpendicular height of the triangle: 12×b×h.

    The question becomes a lot easier if we think about PT as the base of the triangle.

    With PT as the base, the perpendicular height is the vertical line from the x-axis to the point R. In other words, we can get the height from the y-value of R, which is 3 units.

    • PT is a distance of 52(154)=254 units.
    • height is a distance of 3 units.
    Area ΔPRT=12×(254)×(3)=758 units2
    TIP: If we consider PT as the base of the triangle, PR is not the perpendicular height! The perpendicular height must be the line, perpendicular to the base PT, joining the base to the opposite vertex R.

    Submit your answer as:
  2. Calculate, giving reasons, the radius of a circle passing through the points P, T and R.

    INSTRUCTION: Give your answer as a simplified fraction or as a decimal rounded to two places if necessary.
    Answer:

    The diameter of the circle passing through P, R and T is because .

    radius = units.

    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What do you know about a chord that subtends 90° at the circumference of a circle?


    STEP: The radius is always half the diameter
    [−2 points ⇒ 0 / 2 points left]
    A chord that subtends an angle of 90° at the circumference of a circle is a diameter of the circle: chord subtends 90°.

    PT subtends 90° at the circumference of the circle passing through P, T and R.

    PT is the diameter (chord subtends 90°)

    The radius of a circle is always half the diameter.

    r=12×PT=12×254=258 units

    Submit your answer as: andand

ID is: 4402 Seed is: 315

Analytical geometry: area of triangle and circle geometry application

Adapted from DBE Nov 2015 Grade 11, P2, Q4
Maths formulas

In the diagram, not drawn to scale, the straight lines EG and GF are drawn with E(0,1), F(0,173) and G(2,13) such that EGGF.

  1. Calculate the area of ΔEGF.

    INSTRUCTION: Give your answer as a simplified fraction if necessary.
    Answer:

    Area of ΔEGF= units2

    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Consider EF as the base of the triangle. What is the perpendicular height?


    STEP: Use EF and the x-value of G to calculate the area of ΔEGF
    [−3 points ⇒ 0 / 3 points left]
    The area of a triangle is equal to half of its base multiplied by the perpendicular height of the triangle: 12×b×h.

    The question becomes a lot easier if we think about EF as the base of the triangle.

    With EF as the base, the perpendicular height is the horizontal line from the y-axis to the point G. In other words, we can get the height from the x-value of G, which is 2 units. We can't have a negative length! The length of a line segment is always positive, so we take the positive value of 2 units.

    • EF is a distance of 1(173)=203 units.
    • height is a distance of 2 units.
    Area ΔEGF=12×(203)×(2)=203 units2
    TIP: If we consider EF as the base of the triangle, EG is not the perpendicular height! The perpendicular height must be the line, perpendicular to the base EF, joining the base to the opposite vertex G.

    Submit your answer as:
  2. Calculate, giving reasons, the radius of a circle passing through the points E, F and G.

    INSTRUCTION: Give your answer as a simplified fraction or as a decimal rounded to two places if necessary.
    Answer:

    The diameter of the circle passing through E, G and F is because .

    radius = units.

    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    What do you know about a chord that subtends 90° at the circumference of a circle?


    STEP: The radius is always half the diameter
    [−2 points ⇒ 0 / 2 points left]
    A chord that subtends an angle of 90° at the circumference of a circle is a diameter of the circle: chord subtends 90°.

    EF subtends 90° at the circumference of the circle passing through E, F and G.

    EF is the diameter (chord subtends 90°)

    The radius of a circle is always half the diameter.

    r=12×EF=12×203=103 units

    Submit your answer as: andand